'•i-i'i 


.PLANE  TRIGONOMETRY 


AND 


APPLICATIONS 


BY 
E.    J.   WILCZYNSKI,   Ph.D. 

THE    UNIVERSITY   OF   CHICAGO 


EDITED   BY 

H.    E.    SLAUGHT,   Ph.D. 

THE    UNIVEK8ITT   OF   CHICAGO 


ALLYN    AND    BACON 
ISoston  Neto  gorfe  .  Chicago 


,/5^ 


COPYRIGHT,  1914,  BY 
E.  J.  WILCZYNSKI 


PDb 


Nortooolr  ^reaa 

J.  8.  Gushing  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


r- 


PREFACE 

The  characteristic  features  of  this  book  may  be  summarized  as 
follows : 

1.  The  method  of  presentation  is  thoroughly  heuristic.  This  en- 
ables the  student  to  get  a  firm  grasp  of  the  subject  by  teaching 
him  to  recognize  the  fundamental  ideas  which  underlie  and  unify 
the  separate  steps  of  the  mathematical  argument,  instead  of  con- 
fusing him  by  a  disconnected  dogmatic  statement  of  isolated  facts. 

2.  The  book  is  divided  into  ttoo  parts.  The  first  part  is  devoted 
to  the  theoretical  and  numerical  solution  of  triangles.  The 
second  part  treats  of  the  functions  of  the  general  angle,  their 
addition  theorems  and  other  properties,  together  with  applica- 
tions to  simple  harmonic  curves,  simple  harmonic  and  wave 
motion,  and  harmonic  analysis.  Part  One  is  also  published  sepa- 
rately and  is  well  adapted  for  use  in  secondary  schools.  The 
complete  book  is  intended  for  the  freshman  course  in  colleges. 

3.  The  discussion  of  the  solution  of  triangles,  in  Part  One,  is  not 
interrupted  by  any  digressions  about  coordinate  systems,  addition 
theorems,  and  the  like.  It  has  been  thought  desirable  to  post- 
pone to  the  second  part  the  consideration  of  all  of  these  matters, 
which  are  indeed  important  but  unnecessary  for  the  solution  of 
triangles. 

4.  The  definitions  for  the  functions  of  an  obtuse  angle  have  been 
made  to  grow  organically  out  of  the  needs  of  the  problem  of  solv- 
ing triangles,  in  a  way  which  seems  both  simple  and  natural,  and 
which  at  the  same  time  illustrates  an  important  principle  of 
mathematical  procedure. 

5.  The  ivhole  theory  of  triangles  has  been  unified  by  giving  a 
central  position  to  the  area  problem.  As  a  consequence,  almost 
all  of  the  necessary  equations  present  themselves  spontaneously 
and  in  a  connected  fashion.  The  law  of  tangents  is  the  only  one 
which  causes  any  difficulty  in  this  respect.  But  the  law  of  tan- 
gents also  has  been  made  to  submit  to  a  heuristic  treatment,  by 

331458 


iv  PREFACE 

introducing  the  notion  of  the  formrratio  of  a  triangle,  and  combin- 
ing this  notion  with  a  direct  geometric  proof  of  the  formulae  for 
sin  J.  -f  sin  5  and  sin  A  —  sin  B. 

6.  Tlie  numerical  aspect  of  the  work  has  been  discussed  very  fidly. 
Directions  for  computation  are  given  in  great  detail ;  most  of  the 
common  sources  of  error  are  pointed  out ;  and  methods  for  detect- 
ing and  correcting  them  are  indicated.  After  a  thorough  discus- 
sion of  the  significance  of  the  number  of  decimal  places  needed  in 
a  computation,  the  student  is  urged  to  train  and  use  his  judgment 
on  this  matter.  He  is  given  an  opportunity  to  do  this  by  supply- 
ing him  with  complete  five-  and  three-place  tables  and  a  partial 
set  of  four-place  tables. 

7.  The  slide  rule  is  explained  with  considerable  detail  and  its 
use  recommended.  A  number  of  other  labor-saving  devices  are 
discussed. 

8.  The  examples  have  been  selected  with  great  care.  Examples 
without  real  significance  have  been  avoided,  and  the  numbers 
have  been  chosen  so  as  not  to  lead  to  five-place  calculations  when 
such  a  show  of  accuracy  would  be  absurd.  Special  efforts  have 
been  made  to  word  the  examples  in  such  a  way  as  to  avoid 
ambiguity. 

9.  The  applications  cover  a  ivider  field  than  usual,  and  include 
problems  in  heights  and  distances,  surveying,  navigation,  engi- 
neering, astronomy,  and  physics.  But  the  examples  involving 
such  applications  are  not,  as  in  most  texts,  introduced  at  random 
and  without  previous  explanation.  Every  notion  which  is  re- 
quired for.  the  solution  of  any  example  in  the  book  is  fully 
explained  on  the  spot  or  in  some  earlier  portion  of  the  text. 

10.  TJie  use  of  a  few  new  terms,  such  as  the  standard  position  of 
an  angle,  odd  and  even  cardinal  angles,  has  helped  to  simplify 
materially  the  statement  of  a  number  of  important  results. 

11.  The  addition  formulas,  are  presented  in  two  different  ways. 
The  first,  more  elementary  method,  is  made  to  yield  the  general 
result  by  the  help  of  mathematical  induction.  The  second 
method,  based  on  the  notions  of  directed  lines,  line-segments,  and 
angles,  appears  here  in  a  very  simple  and  elegant  form. 

12.  The  articles  on  harmonic  and  laave  motion  tend  to  show  the 
student  that  Trigonometry  has  other  applications  besides  the 
solution  of  triangles. 


r 

PREFACE  V 

13.  A  considerable  amouyit  ofli  istorical  matter  has  been  introduced^ 
not  in  the  form  of  detached  historical  notes,  but  organically  con- 
nected with  the  topic  under  discussion.  Most  of  this  matter  was 
gathered  from  Braunmuhl's  Vorlesungen  ilber  die  Geschichte  der 
Trigonometrie.  Professors  Cajori  and  Karpinski.  have  kindly 
answered  some  questions  of  a  historical  nature  about  which  we 
were  in  doubt. 

14.  The  type  and  the  manner  of  spacing  used  in  the  tables  are 
the  results  of  a  number  of  experiments,  the  object  being  to  pro- 
duce a  set  of  tables  which  should  be  as  pleasant  to  the  eye  as 
possible.  The  tables  are  bound  separately  for  various  reasons. 
In  order  to  make  them  easily  legible,  a  certain  size  of  page  was 
necessary,  and  it  was  thought  undesirable  to  use  so  large  a  page 
for  the  text  itself.  In  the  second  place,  it  is  a  great  advantage 
for  the  student  if  he  can  have  his  text  and  his  tables  open  before 
him  at  the  same  time.  In  the  third  place,  it  is  often  desirable, 
in  examinations,  to  allow  students  to  use  their  tables  without 
their  books.  Pinally,  a  separation  of  the  tables  and  text  makes 
it  easy  to  use  this  text  with  other  tables,  or  these  tables  with 
other  texts,  thus  providing  a  maximum  of  elasticity  in  organizing 
a  course. 

Many  of  the  older  texts  on  Trigonometry  have  been  consulted 
during  the  preparation  of  this  book,  and  the  attempt  has  been 
made  to  learn  from  all  of  them.  The  works  of  Serret,  Lijbsen, 
WiEGAND,  Crockett,  Moritz,  Hall  and  Frink  have  been  espe- 
cially helpful.  A  few  purely  numerical  examples  have  actually 
been  taken  from  these  and  other  texts  without  change,  so  as  to 
reduce  somewhat  the  task  of  computing  the  answers,  and  at  the 
same  time  to  make  the  answers  more  trustworthy.  Most  of  the 
examples,  however,  are  new ;  many  of  them  are  new  in  kind. 

In  conclusion,  the  author  and  editor  wish  to  acknowledge  their 
indebtedness  to  their  colleagues  at  the  University  of  Chicago  for 
various  helpful  suggestions  and  criticisms. 

E.  J.  WILCZYNSKI. 

H.  E.  SLAUGHT,  Editor. 


CONTENTS 

PART  ONE.     SOLUTION  OF  TRIANGLES 

CHAPTER  I 
The  Object  of  Trigonometry. 

PAGB 

1.  Direct  measurement  of  lines 1 

2.  Direct  measurement  of  angles 3 

3.  The  impossibility  of  finding  all  distances  by  direct  measure- 

ment  6 

4.  The  graphic  method 8 

5.  The  desirability  of  an  arithmetical  method  for  solving  triangles        9 


CHAPTER  II 

The  Trigonometric  Functions  of  Acute  Angles. 

6.  The  necessity  of  introducing  new  ideas 11 

7 .  Definition  of  the  trigonometric  functions  of  an  acute  angle      .       13 

8.  The  practical  need  of  tables  giving  the  values  of  the  trigo- 

nometric functions 16 

9.  Relations  between  the  six  trigonometric  functions  of  an  acute 

angle 18 

10.  Relations  between  the  functions  of  complementary  angles        .      20 

11.  The  values  of  the  functions  of  0^,  '60°,  45°,  60°,  90°  .        .        .      22 


CHAPTER  III 

Solution  of  Right  Triangles  by  Natural  Functions. 

12.  Arrangement  and  use  of  the  table  of  natural  functions     .        .       25 

13.  Solution  of  right  triangles  by  means  of  the  table  of  natural 

functions ...       29 

vii 


viii  CONTENTS 

CHAPTER  IV 

PAOB 

Discussion  of  Some  Devices  for  reducing  the  Labor  involved  in  Numer- 
ical Computations. 

14.  The  number  of  decimal  places 34 

15.  The  accuracy  of  a  sum,  difference,  product,  or  quotient  of  two 

numbers  obtained  by  measurement 36 

16.  Labor-saving  devices 37 

17.  Definition  of  logarithms 38 

18.  The  properties  of  logarithms 43 


CHAPTER  V 

Calculation  by  Logarithms. 

19.  Common  logarithms 47 

20.  Properties  of  the  mantissa 48 

21.  Determination  of  the  characteristic  .        .        .        .        .        .  49 

22.  Arrangement  and  use  of  the  table  of  logarithms        ...  51 

23.  Cologarithms     .        . 54 

24.  Extraction  of  roots  by  means  of  logarithms       ....  65 

25.  Logarithmic  calculations  which  involve  negative  numbers        .  56 

26.  The  logarithms  of  the  trigonometric  functions  ....  57 

27.  The  accuracy  of  five-place  tables 59 

28.  The  trigonometric  functions  of  angles  near  0°  or  90°         .         .  60 

29.  The  logarithmic  or  Gunter  scale 61 

30.  The  shde  rule 62 


CHAPTER   VI 

Application  of  Logarithms  to  the  Solution  of  Right  Triangles. 

31.  The  general  method 67 

32.  The  preliminary  graphic  solution 67 

33.  The  gross  errors 68 

34.  Selection  of  formulae  and  checks 68 

35.  The  framework  or  skeleton  form 69 

36.  The  computation 70 

37.  Revision  of  the  computation  when  the  check  is  unsatisfactory  70 

38.  Applications  to  simple  problems  of  surveying,  navigation,  and 

geography 72 

39.  Right  triangles  of  unfavorable  dimensions         .         .         .        .  80 


CONTENTS  ix 

CHAPTER  VII 

PAax 
Theory  of  Oblique  Triangles. 

40.  The  area  of  an  oblique  triangle  in  terms  of  two  of  its  sides  and 

the  included  angle 82 

41.  The  law  of  sines         , 84 

42.  The   law  of    cosines.     A  generalization  of    the  theorem  of 

Pythagoras 85 

43.  Properties  of  the  functions  of  an  obtuse  angle  ....  88 

44.  Other  formulae  for  the  area  of  an  oblique  triangle     ...  90 

45.  The  radius  and  center  of  the  inscribed  circle     ....  93 

46.  The  half-angle  formulae 95 

47.  The  circumscribed  circle .95 

48.  The  form  ratios  of  a  triangle 97 

49.  The  formulae  for  the  sum  and  difference  of  two  sines        .         .  98 

50.  The  law  of  tangents 100 

51.  A  second  proof  of  the  law  of  tangents  and  MoUweide's  equations  102 

CHAPTER  VIII 

Solution  of  Oblique  Triangles. 

52.  The  fundamental  problem 107 

53.  Case  I.   .  Given  one  side  and  two  angles 108 

54.  Case  II.     Given  two  sides  and  the  included  angle     .         .        .  109 
56.    Casein.    Given  two  sides  and  the  angle  opposite  to  one  of  them  111 

56.  Case  IV.     Given  the  three  sides  of  the  triangle         .         .         .116 

57.  Problems  in  heights  and  distances,  plane  surveying,  and  plane 

sailing 118 

58.  Displacements,  velocities,  and  forces 126 

59.  Reflection  and  refraction  of  light 128 


PART  TWO.    PROPERTIES   OF  THE  TRIGONOMETRIC 
FUNCTIONS 

CHAPTER   IX 

The  General  Angle  and  its  Trigonometric  Functions. 

60.  The  notion  of  the  general  angle 133 

61.  Initial  and  terminal  side.     Standard  position  of  an  angle  .  135 

62.  The  notion  of  the  trigonometric  functions  of  a  general  angle    .  136 

63.  Rectangular  coordinates 137 


CONTENTS 

PAGB 

64.  Definition  of  the  trigonometric  functions  of  a  general  angle     .  140 

65.  Discussion  of  the  exceptional  cases 141 

66.  The  four  quadrants 145 

67.  General    character    of    the    trigonometric    functions.      Their 

periodicity 146 

68.  Relations  between  the  trigonometric  functions  of  a  general 

angle 148 

69.  Trigonometric  identities  which  involve  functions  of  a  single 

angle 149 


CHAPTER  X 

Graphic  Representations  of  the  Trigonometric  Functions. 

70.  Line  representation  of  the  trigonometric  functions   .         .         .     151 

71.  Graphs  of  functions,  a  number  of  whose  numerical  values  are 

given 155 

72.  Graphs  of  simple  algebraic  functions         .        .        .        .        .158 

73.  Graphs  of  the  trigonometric  functions 159 

74.  The  natural  unit  of  circular  measurement.    Definition  of  a 

radian 163 

75.  Relations  between  the  functions  of  two  symmetrical  angles      .     167 

76.  Relations  between  functions  of  two  angles  whose  sum  or  differ- 

ence is  a  right  angle 169 

77.  The  quadrantal  formulae 171 

78.  Properties  of  the  sine  and  cosine  curves 175 


CHAPTER  XI 

Relations  between  the  Functions  of  More  than  One  Angle. 

79.  The  addition  theorems  for  sine  and  cosine         .         .        .        .  180 

80.  The  addition  theorems  for  tangent  and  cotangent     .        .        .  185 

81.  The  subtraction  formulae 186 

82.  Formulae  for  converting  products  of  trigonometric  functions 

into  sums,  and  vice  versa 187 

83.  Functions  of  double  angles    - 189 

84.  Functions  of  half  angles 190 

85  a.   The  Umit  ^^^  and  related  limits 198 

d 

85  h.   The  auxiliary  quantities  S  and  T 198 


CONTENTS  XI 
CHAPTER  XII 

PAOB 

Directed  Lines  and  Directed  Line-segments. 

86.  Plan  of  another  proof  for  the  addition  formnlse       .        .        .  201 

87.  Directed  lines  and  segments 201 

88.  Angles  between  directed  lines .         .        .      " .        .         .         .  204 

89.  Projections 204 

90.  Projection  of  a  broken  line 207 

91.  Direction  cosines  of  a  line 209 

92.  Formula  for  the  cosine  of  the  angle  between  two  lines  whose 

direction  cosines  are  given 210 

93.  New  proof  for  the  addition  and  subtraction  formulae       .        .  212 

94.  The  generalized  law  of  sines 212 


CHAPTER  XIII 

The  Inverse  Trigonometric  Functions  and  Trigonometric  Equations. 

95.  The  problem  of  inverting  the  trigonometric  functions     .         .214 

96.  Determination  of  all  of  the  angles  which  correspond  to  a  given 

value  of  one  of  the  functions 214 

97.  The  inverse  trigonometric  functions         .        .         ...         .  217 

98.  Trigonometric  equations 222 

99.  The  equation  a  sin  ^  +  6  cos  ^  =  c 223 


CHAPTER   XIV  w 

Applications  to  the  Theory  of  Wave  Motion. 

100.  Simple  harmonic  motion 226 

101.  The  period  and  phase  constant 228 

102.  Some  illustrations  of  simple  harmonic  motion          .         .         .  229 

103.  Simple  harmonic  curves 231 

104.  Amplitude 232 

105.  Wavelength 232 

106.  Phase  constant 233 

107.  Wave  motion 235 

108.  General  harmonic  motion         .        .         .         .        .         .         .  239 

109.  General  harmonic  curves 241 

110.  Harmonic  analysis  or  trigonometric  interpolation    .         .         .  243 

111.  Theorems  leading  to  the  general  solution  of  the  problem  of 

trigonometric  interpolation 249 

112.  General  solution  of  the  problem  of  trigonometric  interpolation  257 


PLANE  TRIGONOMETRY 

AND 

APPLICATIONS 

PART   ONE 

SOLUTION   OF   TRIANGLES 
CHAPTER   I 

THE   OBJECT   OF  TRIGONOMETRY 

1.  Direct  measurement  of  lines.  One  of  the  most  com- 
mon operations  of  practical  geometry  is  that  of  measuring 
the  distance  between  two  points.  In  its  simplest  form  this 
consists  merely  in  the  repeated  application  of  some  unit  of 
measurement  to  the  required  distance. 

The  units  of  measurement  most  frequently  used  for  this  purpose 
are  a  foot  rule,  a  yardstick,  a  surveyor's  chain,  tape  lines  of  definite 
length,  etc.  Fractional  parts  of  the  unit  are  usually  read  from  a 
graduated  sca/e,  engraved  or  stamped  on  the  standard  used.  A  familiar 
illustration  of  this  is  the  scale  of  inches  on  an  ordinary  foot  rule. 

In  spite  of  the  apparent  simplicity  of  this  process,  it  is  a 
matter  of  great  practical  difficulty  to  carry  out  such  meas- 
urements with  a  high  degree  of  precision.  The  sources  of 
error  are  numerous  and,  in  part,  unavoidable.  No  instru- 
ment made  by  man  is  absolutely  accurate.  Thus,  if  we  use 
a  yardstick,  it  will  not  be  absolutely  straight,  and  it  may  be 
a  trifle  too  long  or  too  short.  It  will  be  very  difficult  to 
make  sure  that  we  are  lajring  off  the  second  yard  of  our  dis- 
tance exactly  where  the  first  yard  ends.  Consecutive  posi- 
tions of  the  yardstick  will  form  angles  with  each  other, 
which  are  not  exactly  equal  to  180°.  In  fact,  it  is  almost 
impossible  to  run  a  straight  line  of  considerable  length,  with 

1 


2  .-.   \  ;  , ;  .      .  THE   OBJECT   OF   TRIGONOMETRY 

any  degree  of  accuracy,  without  the  help  of  more  complicated 
instruments,  such  as  the  transit  described  in  Art.  2. 

The  graduated  scales,  used  for  measuring  fractional  parts 
of  the  unit  of  length,  are  also  affected  by  various  sources  of 
inaccuracy,  and  it  will  be  difficult  for  the  observer  to  esti- 
mate accurately  a  fractional  part  of  the  smallest  visible  divi- 
sion on  the  scale. 

Enough  has  been  said  to  indicate  just  a  few  of  the  many 
difficulties  encountered  in  the,  apparently  so  simple,  opera- 
tion of  measuring  the  length  of  a  line,  and  to  emphasize  the 
fact  that  we  must  always  regard  the  result  of  such  a  meas- 
urement as  an  approximation^  even  if  the  most  refined  instru- 
ments known  to  Science  have  been  used. 

The  difference  between  rough  and  fine  measurements  is  one  of  degree 
only.  The  more  refined  the  method,  the  smaller  will  be  the  "  probable 
error  "  and  the  closer  the  approach  to  the  truth.  But  we  can  never  be 
sure  that  a  quantity  has  been  measured  with  absolute  precision. 

EXERCISE  I 

1.  What  are  some  of  the  sources  of  inaccuracy  in  measuring  the 
length  of  a  table  ? 

2.  If  you  wish  to  measure  the  distance  diagonally  across  a  table,  by 
means  of  a  foot  rule,  what  additional  sources  of  inaccuracy  will  appear  ? 
Would  a  stretched  cord  be  of  some  use  in  this  connection  ? 

3.  How  would  you  measure  the  distance  diagonally  across  a  room 
from  one  of  the  floor  corners  to  the  opposite  corner  of  the  ceiling  ?  Do 
you  know  of  any  other  method  by  which  this  distance  might  be  obtained, 
except  that  of  direct  measurement  ? 

4.  How  would  you  join  two  given  points  by  a  straight  line  (say  for 
the  purpose  of  constructing  a  fence),  over  a  level  piece  of  ground,  if  the 
distance  is  too  great  to  enable  you  to  stretch  a  cord  ?  Do  you  know  of 
any  property  of  the  line  of  sight  which  might  help  in  the  solution  of 
such  a  problem  ? 

5.  If  you  attempt  to  measure  two  different  distances,  of  which  one 
is  about  ten  times  as  great  as  the  other,  using  the  same  foot  rule  and  the 
same  method  of  measurement  in  both  cases,  which  of  these  two  dis- 
tances will  probably  be  obtained  with  greater  accuracy  ?     Why  ? 

6.  What  difficulties  arise,  and  how  would  you  attempt  to  meet  them, 
if  you  were  asked  to  measure  the  horizontal  distance  between  two  points 
on  an  uneven  piece  of  ground  ? 


DIRECT  MEASUREMENT  OF  ANGLES 


7.  Suppose  you  have  measured  a  distance  (say  the  edge  of  a  table) 
with  great  care  and  have  found  it  to  be  equal  to  4  feet  and  9|  inches. 
Is  this  the  exact  length  of  the  table,  or  may  it  be  a  small  fraction  of  an 
inch  greater  or  less  ? 

8.  If  you  were  to  repeat  the  measurement  with  still  greater  care, 
making  use  of  a  more  perfect  standard  of  length,  is  it  likely  that  you 
would  find  exactly  the  same  result  as  before  ? 

9.  In  any  such  measurement  can  you  ever  attain  absolute  accuracy  ? 
If  not,  why  not  ? 

10.  Is  there  any  way  of  knowing  whether  a  measurement  is  absolutely 
accurate  ?  Is  there  any  way  of  knowing  whether  it  is  accurate  to  within 
a  certain  limit  of  accuracy,  say  one  tenth  of  an  inch  ? 

11.  If  a  distance  has  been  measured  by  a  process  which  may  be  relied 
upon  to  give  a  result  accurate  only  to  one  one-hundredth  of  an  inch,  is 
it  desirable,  proper  or  honest,  to  give  the  result  expressed  in  decimal 
parts  of  an  inch  to  more  than  two  decimal  places  ?  Why  not  ?  In  per- 
forming calculations  based  upon  such  measurements,  how  many  decimal 
places  should  we  ordinarily  carry? 

12.  In  measuring  distances  by  means  of  metal  rods,  when  great  ac- 
curacy is  required,  changes  of  temperature  must  be  taken  into  account. 
Why? 

13.  With  what  units  of  length  are  you  familiar  ? 

14.  Gather  from  an  encyclopedia  what  you  can  concerning  the 
"  standard  yard  "  kept  at  Washington. 

15.  What  is  the  metric  system?  What  are  the  relations  to  each 
other  of  the  units  called  millimeter,  centimeter,  decimeter,  and  meter? 
What  is  the  length  of  a  meter  in  inches? 

16.  Can  you  see  any  reason  why  the  metric  system  should  be  prefer- 
able to  the  English  system  of  weights  and  measures  ? 

2.  Direct  measurement  of  angles.  The  operation  of  measur- 
ing angles  is  scarcely  less  important  than  that  of  measuring 
distances.  A  pro- 
tractor is  an  instru- 
ment used  for  this 
purpose.  In  its 
simplest  form,  a  pro- 
tractor consists  of 
an  arc  of  a  circle 
graduated     to     de- 

ees  (Fig.  1).     A  Fig.  i 


THE   OBJECT   OF   TRIGONOMETRY 


mere  inspection  of  the  instrument  will  enable  the  student 
to  see  how  angles  may  be  measured  and  constructed  by  means 
of  it. 

The  unit  usually  employed  in  measuring  angles  is  the 
ninetieth  part  of  a  right  angle,  and  is  called  a  degree.  A 
degree  is  divided  into  sixty  equal  parts,  each  of  which  is  called 
a  minute^  and  each  minute  is  subdivided  into  sixty  seconds. 
Thus 

60  seconds  (60'^)  =  one  minute  (l^? 

60  minutes  (60')  =  one  degree  (1°), 
90  degrees  (90°)  =  one  right  angle. 

Very  frequently,  the  angles  smaller  than  one  degree  are 
expressed  as  decimal  parts  of  a  degree  instead  of  in  minutes 

and  seconds. 

For  the  purpose  of 
measuring  angles  in 
the  field,  surveyors 
make  use  of  an  in- 
strument called  a 
transit  or  theodolite. 
The  essential  parts 
of  this  instrument 
are  (cf.  Fig.  2): 

1.  a  horizontal 
graduated  circle ; 

2.  a  movable  cir- 
cular plate  adjusted 
so  as  to  be  capable  of 
rotation  around  the 
center  of  the  hori- 
zontal graduated  cir- 
cle ; 

3.  an  index  at- 
tached to  the  movable 

v3lk  pla-te  in  such  a  way 

FiQ,  2  as  to  enable  the  ob- 


DIRECT  MEASUREMENT  OF  ANGLES  6 

server  to  read  off  the  amount  of  its  rotation  with  reference  to 
the  fixed  horizontal  circle  ; 

4.  two  standards  attached  to  the  movable  plate  and  carry- 
ing a  horizontal  axis  to  which  is  attached  a  telescope  and 
also,  in  a  complete  instrument,  a  vertical  graduated  circle, 
used  for  measuring  angles  whose  sides  lie  in  a  vertical  plane. 

The  transit  is  usually  supported  on  a  tripod.  If  we  wish  to  measure 
the  angle  between  two  horizontal  lines,  the  tripod  is  placed  over  the 
vertex  of  the  angle  and  the  telescope  is  pointed  toward  some  point  on 
one  of  the  sides  of  the  angle.  The  index  will  then  point  to  a  definite 
division  on  the  horizontal  circle.  The  operation  of  ascertaining  the 
division  of  the  circle  toward  which  the  index  is  pointing,  is  known  as 
"  reading  the  circle."  After  reading  the  circle  and  noting  the  result,  the 
telescope  is  directed  toward  a  point  on  the  other  side  of  the  angle.  The 
difference  between  the  two  readings  of  the  horizontal  cii'cle  will  give 
the  magnitude  of  the  angle. 

In  a  similar  way  the  vertical  circle,  which  is  attached  to  the  axis  of 
the  telescope,  makes  possible  the  measurement  of  angles  whose  sides  lie 
in  a  vertical  plane. 

Both  circles  are  usually  graduated  to  whole  degrees.  The  index,  in 
most  instruments,  is  not  a  simple  pointer,  but  a  so-called  vernier,  an  in- 
genious device  which  enables  the  observer  to  determine  the  reading  of 
the  circle  to  within  a  small  fraction  of  a  degree,  even  if  the  circle  is 
graduated  only  to  whole  degrees.  In  the  most  accurate  instruments,  the 
vernier  is  replaced  by  a  reading  microscope. 

It  is  obviously  very  important  that  the  horizontal  circle  be  exactly 
horizontal,  and  that  its  center  be  exactly  over  the  vertex  of  the  angle 
which  is  to  be  measured.  To  help  in  making  these  adjustments,  the 
surveyor  uses  a  spirit  level  and  a  plumb  line. 

Most  transits  are  also  supplied  with  a  compass,  which  enables  the 
observer  to  determine  the  absolute  directions  of  the  lines  which  he  is 
surveying. 

EXERCISE  II 

1.  Use   a  protractor  to   draw  angles  of   10°,  20'',  31°,  47°  30',  67°, 

78°,  86°. 

2.  Draw  five  angles  at  random  and  measure  them  as  accurately  as 
possible  with  your  protractor. 

3.  Draw  a  triangle  at  random,  measure  its  angles  and  find  their 
sum.  What  should  this  sum  be  ?  If  you  have  obtained  a  different  result 
for  the  sum,  what  are  the  reasons  ? 


6  THE   OBJECT   OF  TRIGONOMETRY 

4.  Construct,  out  of  cardboard,  an  instrument  embodying  the  prin- 
ciple of  the  transit,  substituting  for  the  telescope  some  other  method  of 
taking  a  sight. 

5.  Why  is  it  important  that  the  horizontal  circle  of  a  transit  should 
be  truly  horizontal?  How  does  the  spirit  level  enable  us  to  make  it  so? 
Study  the  article  on  the  spirit  level  in  some  encyclopedia  or  in  a  treatise 
on  surveying. 

6.  Study  the  articles  on  vernier,  micrometer,  reading  microscope, 
compass,  in  an  encyclopedia  or  in  some  appropriate  treatise,  and  write  an 
abstract  of  the  same. 

7.  Describe  the  sources  of  inaccuracy  which  you  think  may  arise  in 
the  measurement  of  an  angle  by  means  of  a  protractor  or  theodolite. 

8.  Deduce  rules  for  converting  minutes  and  seconds  into  decimal 
parts  of  a  degree  and  vice  versa. 

,     9.    Apply  this  rule  to  the  angles  ,  \ 

'  30°.  20',  10°  45',  8°  40' 20",  3°'8'V'. 

10.   Express  the  following  angles  in  degrees,  minutes,  and  seconds : 
23°.14,  18°.^5,  46^235. 


3.  The  impossibility  of  finding  all  distances  by  direct 
measurement.  We  have  discussed  briefly  the  direct  methods 
for  measuring  distances  and  angles.  It  is  clear  from  our 
discussion  that  it  is  very  difficult  to  measure  great  distances 
in  that  way.  But  there  are  many  cases  in  which  it  is  al- 
together impossible  to  apply  such  direct  methods.  How,  for 
example,  should  we  proceed  to  find  the  distance  through  a 
mountain  or  across  an  extensive  valley  ?  How  shall  we  find 
the  distance  from  New  York  to  London,  or  from  the  earth  to 
the  moon,  by  direct  measurement  ? 

It  is  clear  that,  if  we  wish  to  answer  such  questions  at  all, 
we  shall  have  to  devise  some  method  different  from  that  of 
direct  measurement. 

The  attempt  to  solve,  by  indirect  methods,  problems  whose  direct 
solution  is  inconvenient  or  impossible  usually  leads  to  great  advances  in 
Science.  The  most  important  theorems  of  elementary  geometry  were 
probably  first  discovered  by  the  Ancients  in  their  attempts  to  devise  con- 
venient and  practical  methods  for  the  measurements  which  they  found 
necessary  for  the  purposes  of  their  everyday  life.  It  is  generally  believed, 
for  instance,  that  the  Egyptians  became  expert  geometricians  and  sur- 
veyors at  an  early  period  of  their  history,  because  it  was  so  important 


DIFFICULTY   OF   DIRECT  MEASUREMENT  7 

for  them  to  be  able  to  reestablish  the  boundary  lines  of  their  lands  after 
the  effacement  of  their  marks  by  the  annual  inundations  of  the  Nile. 

The  earliest  Greek  philosophers  and  mathematicians  were  pupils  of 
the  Egyptians,  and  some  of  their  first  achievements  were  connected  with 
problems  of  the  particular  kind  which  we  are  now  discussing.  Thus  it 
is  reported  that  Thalp:8  of  Miletus  (about  600  B.C.)  measured  the 
height  of  a  pyramid  by  measuring  the  length  of  its  shadow,  at  the  instant 
when  the  shadow  of  a  vertical  stick  by  its  side  was  exactly  as  long  as  the 
stick  itself.  The  height  of  the  pyramid  would  then  be  equal  to  the 
length  of  its  shadow  at  that  moment. 

This  method  has  the  inconvenient  feature  of  compelling  the  observer 
to  wait  (many  hours  perhaps)  for  the  right  moment.  According  to  a 
report  by  Plutarch,  Thales  also  devised  a  second  method  which 
avoided  this  inconvenience.  This  method  involves  the  use  of  the  simplest 
properties  of  similar  triangles  and  may  be  illustrated  by  means  of  the 
following  example : 

We  place  a  stick  5  feet  high  into  the  ground  near  a  building  whose 
height  we  wish  to  find.     At  any  convenient  moment  we  measure  the 


125  feet 


Fig.  3 


length  of  the  shadows  of  the  stick  and  of  the  building.  Suppose  we 
find  in  this  way  that  the  shadow  of  the  building  is  125  feet  long  at  the 
moment  when  the  shadow  of  the  stick  is  10  feet  long.  If  h  denotes  the 
height  of  the  building,  we  shall  have  (Fig.  3) 


whence 


A:  125 
h 


5:10, 
62.5  feet. 


According  to  Eudemus  (about  300  B.C.),  one  of  the  earliest  writers 
on  the  history  of  mathematics,  Thales  also  invented  a  method  for 
measuring  the  distance  from  the  shore  to 
a  ship  at  sea.  Although  Eudemus  does 
not  describe  Thales's  method  in  detail,  he 
says  enough  to  lead  us  to  conclude  that  it 
was  essentially  as  follows  :  Let  5L  be  a 
tower,  say  a  lighthouse  on  the  shore,  and 


8  THE   OBJECT   OF  TRIGONOMETRY 

let  S  be  the  ship  at  sea.     Measure  the  angle  BLS  and  the  height  of  the 
tower.     Construct  a  similar  triangle  B'L'S'  on  the  drawing  board  and 
measure  B'S'  and  B'L'.     Then  BS  can  be  found  from  the  proportion 
BS'.BL  =  B'S'.B'L'. 

4.  The  graphic  method.  The  examples  discussed  in  Art. 
3  show  how  we  may  solve  many  problems  of  practical  geom- 
etry by  indirect  measurements.  We  did  not  measure  directly 
the  quantity  which  we  were  seeking,  but  some  other  related 
quantities,  and  ultimately,  by  means  of  these  relations,  we 
determined  the  desired  quantity  itself. 

But  these  same  examples  may  serve  to  illustrate  another 
point.  The  solutions  which  we  gave  are  examples  of  the 
graphic  method,  that  is,  of  a  process  which  makes  use  of 
drawing  instruments,  of  accurate  geometric  constructions 
and  measurements,  for  the  purpose  of  obtaining  the  values 
of  the  unknown  quantities.  Such  graphical  methods  are 
often  extremely  valuable  and  have  been  developed  in  recent 
times,  in  connection  with  other  parts  of  mathematics,  so  as 
to  give  rise  to  very  important  results. 

The  graphic  solution  of  any  problem  about  triangles  will  finally  re- 
duce to  an  application  of  certain  theorems  of  geometry  which  state  that 
a  triangle  is  determined  and  may  be  constructed  when  certain  ones  of  its 
six  parts  (sides  and  angles)  are  given.  It  is  clear,  then,  that  these 
theorems  are  particularly  important  for  the  graphic  method.  Some  of 
the  following  questions  have  been  chosen  for  the  purpose  of  aiding  the 
student  to  refresh  his  memory  in  regard  to  these  matters. 

EXERCISE  III 

In  the  following  examples  and  throughout  the  book  we  shall  usually 
denote  the  angles  of  a  triangle  hj  A,B,  C  and  the  sides  opposite  to  these 
angles  by  a,  b,  c,  respectively.  The  student  should  be  provided  with  a 
protractor,  a  pair  of  compasses,  and  a  ruler  divided  decimally,  say  into 
centimeters  and  millimeters.  All  constructions  and  measurements  should 
be  made  as  carefully  as  possible. 

1.   Given  a  =  3.72,  b  =  4.91,  c  =  2.56.     Find  the  angles. 
v^   2.   Given  a  =  4.27,  B  =  35°,  C  =  69°.     Find  the  remaining  parts  of 
the  triangle. 

3.  Given  b  =  5.63,  c  =  6.71,  A  =  27°.  Find  the  remaining  parts  of 
the  triangle. 


AN   ARITHMETICAL   METHOD  9 

4.  Given  a  =  4.23,  b  =  5.16,  A  =  55°.  Find  the  remaining  parts  of 
the  triangle. 

5.  When  a,  b,  c  are  given  at  random,  can  we  always  find  a  triangle 
of  which  o,  &,  c  are  the  sides,  or  is  there  some  restriction  on  the  possible 
values  of  a,  &,  c  ? 

6.  If  a,  b,  A  are  given,  there  may  be  one  or  two  solutions,  or  no 
solution.     Discuss  these  cases. 

7.  Is  a  triangle  determined  when  we  know  the  magnitudes  of  its  three 
angles?  Why?  When  the  three  angles  of  a  triangle  are  given,  have  we 
obtained  essentially  more  information  than  if  only  two  of  them  are 
given  ?  Why  ?  Is  the  third  angle  of  a  triangle  independent  of  the  other 
two? 

8.  What  do  you  mean  by  similar  triangles? 

9.  Under  what  conditions  are  two  triangles  similar  ? 

10.  Use  the  shadow  method  of  Thales  (Art.  3)  to  measure  the  height 
of  some  building  in  your  neighborhood. 

11.  By  means  of  a  transit,  or  else  by  means  of  the  home-made  in- 
strument of  cardboard  suggested  in  Ex.  4  of  Exercise  II,  find  the  dis- 
tance of  some  object  situated  on  the  opposite  side  of  the  street  from  your 
home,  without  crossing  the  street.  Afterward  check  your  result  by 
direct  measurement. 

12.  How  may  a  person  on  board  ship  find  his  distance  from  a  build- 
ing on  the  shore  if  he  knows  its  height? 

^j 
5.  The  desirability  of  an  arithmetical  method  for  solving 
triangles.  We  have  seen  that  it  is  a  rather  simple  matter  to 
solve  a  triangle  by  the  graphic  method.  But  we  can  hardly 
feel  altogether  satisfied  with  the  graphic  solution,  for  it  will 
clearly  not  permit  us  to  reach  any  great  degree  of  accuracy. 
To  be  sure,  by  making  our  drawings  on  a  very  large  scale, 
we  might  lay  off  distances  with  considerable  precision,  but 
we  should  still  encounter  the  difficulty  of  accurately  plotting 
angles.  Clearly  it  would  require  extraordinary  skill  and 
exceedingly  fine  instruments  to  enable  us  to  draw  an  angle 
so  accurately  that  its  error  should  not  exceed  one  minute  of 
arc.  Many  other  circumstances  combine  to  make  a  graphical 
solution  unsatisfactory  if  a  high  degree  of  precision  is  required. 

The  main  value  of  the  graphic  method  lies  in  furnishing  a 
solution  whose  approximate  correctness  is  apt  to  he  apparent  to 


10  THE   OBJECT   OF   TRIGONOMETRY 

the  eye^  and  which  may  therefore^  in  almost  all  cases^  serve  as  a 
check  on  the  more  complete  solution  obtained  in  some  other  way. 

But  the  lack  of  accuracy  is  only  one  of  the  defects  of  the 
graphic  method,  although  perhaps  the  most  important  one 
from  a  practical  point  of  view.  The  other  defect  which  we 
wish  to  emphasize  is  more  of  a  theoretical  nature.  The  parts 
of  a  triangle  (its  sides  and  angles)  are  usually  given  as 
numbers  (so  many  feet,  so  many  degrees).  It  seems  natural, 
therefore,  to  suppose  that  there  must  exist  some  arithmetical 
process  for  the  solution  of  triangles. 

Trigonometry  enables  us  to  find  the  unknown  parts  of  a 
triangle  by  arithmetical  processes. 

This  statement  must  not  be  regarded  as  a  complete  defi- 
nition of  trigonometry.  We  shall  see  later  that  the  solution 
of  triangles  by  arithmetical  methods  constitutes  only  a  part, 
although  an  important  part,  of  trigonometry. 

EXERCISE  IV 

1.  What  are  the  principal  sources  of  inaccuracy  in  solving  a  triangle 
by  the  graphic  method  ? 

2.  We  wish  to  construct  an  isosceles  triangle,  given  the  angle  at  the 
vertex  and  the  altitude.  Suppose  that  the  error  made  in  constructing 
the  angle  is  5'.  Will  the  effect  of  this  error  on  the  base  of  the  triangle, 
as  obtained  from  the  construction,  be  different  for  different  altitudes? 
Will  it  be  greater  or  less  for  the  higher  triangles  ? 

3.  If  you  wish  to  find  a  point  as  the  intersection  of  two  straight  lines, 
are  you  likely  to  obtain  a  more  accurate  result  if  the  two  lines  are  at 
right  angles,  or  if  they  are  nearly  parallel? 

4.  A  side  and  two  adjacent  angles  of  a  triangle  are  given,  and  its 
other  parts  are  to  be  found  graphically.  Is  such  a  graphic  solution  likely 
to  be  accurate  if  both  of  the  angles  are  very  small  ?  If  the  sum  of  the 
given  angles  is  very  close  to  180°? 

5.  Formulate,  in  your  own  words,  the  distinction  between  an  arith- 
metical and  a  graphical  process. 


CHAPTER   II 

THE   TRIGONOMETRIC   FUNCTIONS   OF   ACUTE   ANGLES 

6.  The  necessity  of  introducing  new  ideas.  At  the  close 
of  Chapter  I  we  formulated  the  problem  of  devising  arith- 
metical methods  for  the  solution  of  triangles.  But,  if  we 
think  of  the  many  shapes  which  a  triangle  may  assume,  this 
problem  appears  to  be  a  most  formidable  one.  We  shall, 
therefore,  for  the  present,  confine  our  attention  to  the  com- 
paratively simple  case  of  a  right  triangle.  We  have  good 
reason  to  suppose  that,  if  we  succeed  in  solving  our  problem 
for  all  right  triangles,  we  shall  be  able  to  deal 
later  with  the  general  case  also,  since  every 
triangle  may  be  decomposed  into  two  right 
triangles. 

Let  us  then  consider  a  triangle  ABC^  right- 
angled  at   0  (Fig.  5),  and  let  us  denote  the 
sides  opposite  the  angles  A^  B,  C  by  a,  5,  <?,  re-  ^'        6" 
spectively.     We   are   already  acquainted    with         ^^^-  ^ 
two  relations  which  will  assist  us  in  the  solution  of   our 
problem,  namely :  ^ 

(1)  .4  +  ^=90° 

and  the  theorem  of  Pythagoras,  ^^        >^ 

(2)  a^  +  b^=(^.  "^     H^<^^-^ 

The  first  of  these  equations  enables  us  to  calculate  one  of  the 
acute  angles  if  the  other  one  is  given.  The  second  provides 
a  method  for  calculating  any  one  of  the  three  sides  if  the 
other  two  are  given. 

But  we  have  nothing  as  yet  which  will  enable  us  to  find 
the  angle  A  if  two  of  the  sides  (say  a  and  6)  are  given, 
although  the  triangle  is  clearly  determined  by  these  sides 
and  might  be  constructed  by  geometry.     The  equations  (1) 

11 


12         FUNCTIONS  OF  ACUTE  ANGLES 

and  (2),  unaided  by  other  relations,  are  obviously  inadequate 
for  this  purpose,  since  (1)  is  a  relation  between  the  angles 
A  and  B  alone,  while  (2)  involves  only  the  sides  of  the  tri- 
angle. Now,  clearly,  a  statement  which  is  only  concerned 
with  the  angles  of  a  triangle  cannot  convey  any  positive 
information  about  t\\Q  sides ^  and  vice' versa. 

In  order  that  we  may  he  able  to  solve  a  right  triangle  by 
arithmetical  processes^  there  must^  therefore^  be  added  to  equa- 
tions (1)  arid  (2)  certain  other  relations^  in  which  the  sides  and 
angles  shall  not  be  separated,  but  in  which  they  shall  occur 
simultaneously. 

The  discovery  of  such  relations  and  their  adequate  formulation 
is  the  foundation  upon  which  all  of  trigonometry  must  finally 
rest. 

A  simple  illustration  will  make  clear  the  nature  of  these  new  relations. 

The  numerical  measure  for  the  steepness  of  an  inclined  plane,  say  a 

mountain  road,  may  be  given  in  two  ways.     We  may  say  that  the  road 

makes  a  certam  angle  A  (say  5°)  with  the  horizontal  plane,  or  we  may 

say  that  it  rises  a  certain  number  of  feet  (say  87.5  feet)  in  a  horizontal 

distance  of  1000  feet. 

87  5 
The  quotient,  — '—,  is  technically  known  as  the  slope,  grade,  or  gradient 

of  the  road.     The  gradient  is  clearly  connected  with  the  angle  A  in  such  a 

way  that,  if  A  should  vary,  to  every  value 
of  the  angle  there  corresponds  a  definite 
value  of  the  gradient  and  vice  versa.  Thus, 
if  AB  (Fig.  6)  represents  the  road,  and  if 
a  and  b  are  both  expressed  in  feet,  the  gra- 
dient of  AB  is  equal  to  -.     The  value  of  this  quotient  changes  with  the 

b 
angle  A,  so  that  for  different  angles  we  find  different  values  for  the 
gradient. 

The  general  question  before  us  may  be  formulated  as  fol- 
lows. If  the  acute  angle  A  of  the  right  triangle  ABO  be- 
comes larger  or  smaller,  what  effect  will  such  a  change  have 
upon  the  sides  ?  And,  conversely,  if  the  sides  of  a  triangle 
change,  what  effect  will  this  have  on  the  angles  ? 

Now,  since  similar  triangles  have  their  corresponding 
angles  equal,  it  is  clear  that  there  are  some  changes  in  the 


DEFINITIONS  OF  FUNCTIONS  13 

lengths  of  the  sides  which  produce  no  change  in  the  angles. 
In  fact,  if  all  of  the  sides  of  a  triangle  are  magnified  in  the 
same  ratio,  the  angles  are  not  changed 
at  all. 

In  the  two  right  triangles  ABC 
and  A'B'Q'  (Fig.  7)  we  have 

a  _a!     h  _h'     « _  «^     ^ 
c      c'     e      c'     h      h' 

if  the  angle  at  A'  is  equal  to  that  at 
A,     But  if  the  angle  A'  is  different 

from  A,  the  ratios  —,,  -,  ^,  etc.,  will  not  be  equal  to  -,  -,% 
e     G     0  c     c    0 

etc.,  respectively.     For  if  they  were,  corresponding  pairs  of 

sides  of  the  two  triangles  would  have  the  same  ratio,  the 

triangles  would  be  similar,  and,  contrary  to  our  hypothesis, 

angle  A!  would  have  to  be  equal  to  angle  A. 

Consequently,  while  the  lengths  of  the  indivicbual  sides  of 

a  right  triangle  have  nothing  to  do  with  the  size  of  its  angles, 

the  ratios  of  these  lengths  are  connected  with  the  magnitude 

of  the  angles  in  a  very  intimate  fashion.     In  fact,  so  close  is 

this  relation  that  the  values  of  the  ratios  -,  -,  -,  etc.,  may  be 

G     c    0 

determined  (by  construction)  as  soon  as  the  acute  angle  A 
has  been  chosen,  and  conversely ;  if  one  of  these  ratios  is 
given,  we  can  find  (by  construction)  one,  and  only  one,  cor- 
responding acute  angle  A. 

7.    Definitions  of  the  trigonometric  functions  of  an  acute 

angle.     We  have  seen  that  the  values  of  the  ratios  -,  -,  etc., 

c    c 

of  the  sides  of  a  right  triangle  are  closely  bound  up  with  the 
magnitude  of  the  angle  A.  If  the  angle  changes,  each  of 
these  ratios  changes  and  vice  versa. 

Now,  one  variable  quantity  is  called  a  function  of  another^  if 
they  are  so  related  that  any  change  in  the  latter  produces  a 
corresponding  change  in  the  former. 


14  FUNCTIONS  OF  ACUTE  ANGLES 

Consequently,  each  of  the  six  ratios  -,-,-,-,-,-,  deter- 

c  c  h  a  a  h 
mined  by  the  sides  of  the  right  triangle  ABQ^  is  -a,  function 
of  the  angle  A^  because  any  change  in  A  produces  a  corre- 
sponding change  in  each  of  those  ratios.  Each  of  these 
functions  has  received  a  name  and  a  symbol.  The  reason 
for  choosing  these  names  will  appear  later  (see  Arts.  10  and 
70),  and  cannot  be  discussed  with  profit  at  the  present 
moment. 

We  proceed  to  give  the  formal  definitions  of 
the  six  trigonometric  functions  of  an  acute 
angle  A. 


Construct  any  right  triangle  (cf.  the  Fig.),  one 

of  whose  acute  angles  is  equal  to  the  given  angle 

A.*     Of  the  two  legs  of  this  right  triangle^  that 

one  which  passes  through  the  vertex  of  the  angle 

A  is  said  to  he  adjacent  to  A.     The  other  leg  is  said  to  he 

opposite  to  A,  and  the  third  side  of  the  triangle  is  called  its 

hypotenuse. 

The  sine  of  A   is   the  ratio   of   the  opposite  side  to  the 
hypotenuse. 

The  cosine  of  A  is  the  ratio  of  the  adjacent  side  to  the 
hypotenuse. 

The  tangent  of  A  is  the  ratio  of  the  opposite  side  to  the 
adjacent  side. 

The  cotangent  of  A  is  the  ratio  of  the  adjacent  side  to  the 
opposite  side. 

The  secant  of  A  is  the  ratio  of  the  hypotenuse  to  the  ad- 
jacent side. 

The  cosecant  of  A  is  the  ratio  of  the  hypotenuse  to  the 
opposite  side. 


*  We  have  seen  in  Art.  6  that  the  size  of  this  right  triangle  is  absolutely  o-f 
no  consequence,  since  any  two  triangles  of  this  kind  are  similar,  so  that  the 
corresponding  ratios  for  the  two  triangles  will  be  equal. 


DEFINITIONS  OF  FUNCTIONS  15 

In  symbols  we  may  write  these  definitions  as  follows : 
sin  ^  =  - ,       cos  ^  =  - ,       tan  ^  =  ^  ^ 

(1)         :       :       I 

cseA  =  —  ,       sec  A  =  ^f        cotA  =  —» 

These  symbols  are  written  abbreviations  of  the  names  of  the  functions. 
In  speaking,  the  symbols  are  pronounced  as  though  the  name  of  the  func- 
tion had  been  written  out  in  full.  Thus,  tan  A  is  pronounced  tangent 
of  A  or  tangent  A  ;  esc  A  is  pronounced  cosecant  of  A  or  cosecant  A,  etc. 

In  defining  the  trigonometric  functions,  we  made  use 
of  the  numerical  measures  of  certain  line-segments,  namely,  of 
the  sides  of  a  right  triangle.  Now,  the  numerical  measure 
of  a  line-segment  changes  if  the  unit  of  measurement  is 
changed.  One  might,  therefore,  expect  the  values  of  the 
trigonometric  functions  of  an  acute  angle  to  change  with 
every  change  of  the  unit  of  length.  But  this  is  not  the  case. 
Owing  to  the  fact  that  only  ratios  of  these  line-segments 
appear  in  equations  (1),  the  functions  sin  A^  cos  A^  etc.,  are 
found  to  have  the  same  value  whether  the  line-segments  a,  6, 
c  are  measured  in  feet,  inches,  or  in  terms  of  any  other  unit 
of  length. 

Consider,  for  instance,  a  right  triangle  for  which  a  =  3  feet,  &  =  4 
feet,  c  =  5  feet.  According  to  (1)  we  have  sin  ^  =  f ,  cos  A  =  f,  etc. 
Let  us  now  introduce  the  inch  as  unit  of  length  instead  of  the  foot.  The 
numerical  measures  of  the  sides  of  the  triangle  will  now  be  a  =  36, 
b  =  4:8,  c  =  60.  According  to  (1)  we  shall  now  find  sin  A  =  |^, 
cos  A  =  If,  etc.  But  |§  =  f ,  |f  =  f,  etc.,  so  that  we  obtain  precisely 
the  same  values  for  sin  A,  cos  A,  etc.,  whether  the  sides  of  the  triangle 
be  expressed  in  feet  or  inches. 

If  the  trigonometric  functions  were  concrete  numbers,  that 
is,  if  they  were  the  numerical  measures  of  some  kind  of  con- 
crete quantity  such  as  a  length,  an  area,  or  a  volume,  their 
values  would  change  every  time  that  a  change  is  made  from 
one  unit  of  length  to  another.  We  have  just  seen  that  this 
is  not  the  case.  Therefore,  the  trigonometric  functions  are 
pure  or  abstract  numbers.* 

*  This  same  fact  is  sometimes  (somewhat  inadequately)  expressed  by  the 
statement  that  the  trigonometric  functions  are  ratios. 


16         FUNCTIONS  OF  ACUTE  ANGLES 

8.  The  practical  need  of  tables  giving  the  values  of  the 
trigonometric  functions.  The  trigonometric  functions  just 
defined  will  enable  us  to  find  the  unknown  parts  of  a  right 
triangle  when  certain  parts  are  given,  provided  only  that  we 
can  devise  a  practical  method  for  actually  obtaining  the 
numerical  values  of  these  functions  for  any  given  acute 
angle.  Now,  the  values  of  the  functions  have  been  calcu- 
lated by  mathematicians  and  the  results  have  been  collected 
in  tabular  form  for  convenient  use.  From  the  practical  point 
of  view,  therefore,  it  only  remains  to  explain  the  arrange- 
ment and  the  use  of  the  tables. 

To  the  more  scientifically  inclined  student,  however,  the 
question  will  immediately  suggest  itself  as  to  how  these  use- 
ful tables  were  actually  computed.  We  shall  reserve  the 
answer  to  this  question  for  the  second  part  of  the  book. 
The  following  examples,  however,  will  show  how  a  table  of 
trigonometric  functions  may  be  prepared  by  the  graphic 
method,  provided  that  no  very  high  degree  of  accuracy  be 
required. 

EXERCISE  V 

1.   Find  the  functions  of  40°  by  the  graphic  method. 
Solution.    With  the  help  of  a  protractor  construct  the  angle  PAQ 
(Fig.  8)  equal  to  40°.     Choose  any  point  B  on  AQ  and  drop  a  perpendic- 
ular BC  from  5  to  ^P.     Measure  the  distances 
BC,  AC,  and  AB.     Then  will 

(1)  sin 40°=^,  cos40°=4i- 

AB  AB 

.P         Although  the  point  B  might  be  chosen  anywhere 

on  AQ,  it  will  be  especially  convenient  to  make  AB 

equal  either  to  one  unit,  ten  units,  or  one  hundred 

units.     For,  as  equations  (1)  show,  we  have  to  divide  by  AB,  and  if 

AB  is  equal  to  1,  10,  or  100,  we  avoid  the  long  division  which  would 

otherwise  be  necessary. 

Let  us,  therefore,  make  AB  =  10  centimeters.     We  should  then  find 

by  measurement,        ^^      a  a  a  r^      t^t 

•^  '        C5  =  6.4  cm.,  ^C  =  7.7  cm. 

According  to  (1),  therefore, 

sin  40°  =M  ^  0.64,  cos 40°  =  "^  =  0.77. 
10  10 


VALUES  OF  FUNCTIONS  17 

Further  we  find 

tan  40°  =  —  =  0.83,   cot  40°  =  ^  =  1.20, 

sec  40°  =  —  =  1.30,   CSC  40°  =—  =  1.56. 

7.7  6.4 

We  should  obtain  a  more  accurate  result  for  tan  40°,  and  more  con- 
veniently, if  we  were  to  use  another  triangle,  making  this  time  ^C  =  10 
cm.     Measurement  would  then  give  BC  =  8.4  cm.,  and 

tan  40°=— =0.84. 
10 

2.  Find  the  functions  of  the  following  angles  by  the  graphic  method : 

(a)  10°.     (6)  15°.     (c)  20°.     {d)  70°. 
Construct  carefully  each  of  the  following  right  triangles,  measure  the 
angles,  and  find  the  six  functions  of  each  acute  angle. 

3.  a  =  3,  6  =  4,  c  =  5. 

4.  a  =  5,   6  =  12,   c  =  13.    ^ 

5.  a  =  8,   &  =  15,   c  =  17.  /;,  ,. 

6.  Construct  and  measure  an  acute  angle  whose  sine  is  equal  to  \. 

7.  Construct  and  measure  an  acute  angle  whose  tangent  is  equal  to  f. 

8.  Construct  and  measure  an  acute  angle  whose  cosine  is  equal  to  |. 

9.  Can   you  think  of  two  right  triangles  (Fig.  7)    with   different 
angles  A  and  A',  for  which  the  sides  a  and  a'  are  nevertheless  equal? 

10.  Can  you  conceive  of  two  right  triangles  (Fig.  7)  with  different 

angles  A  and  A'y  for  which  the  ratios  -  and  —  are  nevertheless  equal? 

b  b' 

11.  Why,  then,  do  we  speak  of  the  ratio  ~  as  a  function  oi  Al  Why 
do  we  not  introduce  a  or  db  as  a  function  of  yl  ? 

12.  Assuming  that  we  have  access  to  a  table  of  the  values  of  the 
trigonometric  functions,  show  how  we  might  solve  each  of  the  following 
problems.  To  find  the  remaining  parts  of  a  right  triangle  when  the 
following  parts  are  given. 

I.  a,  &,  C  =  90°.  Til.  a,  c,  C  =  90°. 

II.  a.  A,  C  =  90°.  IV.   c,A,C=  90°. 

13.  Show  that  neither  the  sine  nor  the  cosine  of  an  acute  angle  can 
ever  be  greater  than  unity. 

14.  Show  that  the  tangent  of  an  acute  angle  may  have  any  positive 
value  whatever.     Similarly  for  the  cotangent. 

15.  What  restrictions,  if  any,  are  there  on  the  values  which  the  secant 
and  cosecant  of  an  acute  angle  may  assume  ? 


18 


FUNCTIONS  OF  ACUTE  ANGLES 


9.  Relations  between  the  six  trigonometric  functions  of  an 
acute  angle.  The  preceding  discussion  suffices  to  indicate 
the  importance  of  constructing  a  table  of  the  values  of  the 
trigonometric  functions.  The  task  of  computing  these  tables 
may  be  abbreviated  very  considerably  by  noting  that  the  six 
functions  are  not  independent  of  each  other.  In  fact,  we 
have  (Fig.  9) 


(1) 


sin^ 


cos^  = 


tan  J. 


CSC  J.= 

sec  J.  = 
cot^  = 


so  that  we  obtain  at  once  the  relations 


csc^  = 


sin  J. 


sinvl 
,1 


sec  J.  = 


cos  J.  = 


sec^' 


cot  A' 


or, 
(2)       sm^csc^  =  l,    cos^sec^  =  l,    tan^cot^  =  1.* 

But,  two  numbers  whose  product  is  equal  to  unity  are  called 
reciprocals  of  each  other.  Therefore,  equations  (2)  are 
equivalent  to  the  following  statement : 

The  sine  and  cosecant^  the  cosine  arid  secant,  and  finally/  the 
tangent  and  cotangent,  of  an  acute  angle  are  reciprocals. 

Clearly,  knowledge  of  this  fact  reduces  greatly  the  labor  of  computing 
tables  of  the  functions.  For,  having  found  the  values  of  the  three 
functions,  sine,  cosine,  and  tangent,  the  values  of  the  remaining  three  can 
be  obtained  from  these  by  computing  their  reciprocals. 

But  there  are  other  relations  besides' (2)  which  enable  us 
to  reduce  still  further  the  labor  involved  in  constructing, a 
table  of  the  trigonometric  functions.     We  have 

tan^  =  f. 

0 


*  Note  that  sin  A  esc  A  is  written  for  sin  4  x  esc  -4  just  as  ab  is  written  for 
ax  6. 


RELATIONS  BETWEEN   FUNCTIONS  19 

If  we  divide  both  numerator  and  denominator  of  the  fraction 
-  by  c,  we  find 

b/c 
Bat  we  have,  by  definition  (cf.  equations  (1)) 

-  =  sin  ^,  -  =  cos  A^ 
c  c 

and  therefore 

(3)  tan^  =  ^i^.  « 

Of  course,  since  cot  A  is  the  reciprocal  of  tan  A^  we  also  have 

The  relations  (2),  (3),  (4)  enable  us  to  calculate  the  values 
of  all  six  functions  when  the  sine  and  cosine  are  known. 
But  it  actually  suffices  to  know  the  sine.  For,  if  we  divide 
both  members  of  the  familiar  equation 

(5)  a2  +  62  =  c2 

by  c2,  we  find 


But,  by  definition,  we  have 

a       •     A     h  A 

—  =  sin  A^   -  =  cos  A^ 
c  c 

so  that  (5)  becomes 

(6)  sm2^  +  cos2^=  1, 

where  sin^  J.  and  go^^  A  have  been  written,  as  is  customary, 
for  (sin  J.)^  and  (cos  J.)^,  respectively. 

It  follows  from  relations  (2),  (3),  (4),  and  (6)  that,  if  we 
know  the  value  of  a  single  one  of  the  six  trigonometric 
functions  of  an  acute  angle,  the  values  of  the  remaining  five 
may  be  computed.  The  detailed  proof  of  this  statement  will 
be  left  to  the  student  in  some  of  the  examples  given  below. 


■s 


20  FUNCTION'S   OF   ACUTE  ANGLES 

EXERCISE  VI 

In  each  of  the  following  twelve  examples,  the  value  of  one  function  of 
the  acute  angle  A  is  given.     Find  the  values  of  the  remaining  functions. 

1.  sin^  =  |.  4.   cot  ^  =  2.  7.   sin  ^  =  x,  10.   cot^=a:. 

2.  cos^=T^3.  5,   sec^  =  ^f.  8.   cos^=a:.  11.   sec^=a:. 

3.  tah^  =  l.  6.   cscA=^^.  9.   tan^=a:.  12.   csc^=af. 

13.   Construct  an  isosceles  right  triangle  and  make  use  of  this  figure 
for  the  purpose  of  computing  the  functions  of  45°. 

■y  1^  Divide  an  equilateral  triangle  into  two  right  triangles  by  dropping 
a  perpendicular  from  one  of  its  vertices  to  the  opposite  side.  Make  use 
of  this  figure  for  the  purpose  of  computing  the  functions  of  30°  and  60°. 

15.  Collect  in  a  table  the  results  of  Exs.  13  and  14. 

16.  Prove  the  formula  sec^  A  =  l-\-  tan^  A. 

17.  Prove  the  formula  csc^  A  =  l  +  cof^  A.' 

18.  Prove  that  the  sine,  tangent,  and  secant  of  an  angle  increase  when 
the  angle  grows  from  0°  to  90°. 

19.  Prove  that  the  cosine,  cotangent,  and  cosecant  of  an  angle  decrease 
when  the  angle  grows  from  0°  to  90°. 

20.  Prove  that  tan  ^  <  1,  cot  ^  >  1  if  J  <  45°,  and  that  tan  A  >  1, 
cot^<lif  ^  >45°. 

21.  Show  that  each  of  the  six  functions  may  be  expressed  either  as  a 
product  or  as  a  quotient  of  two  of  the  others. 

10.  Relations  between  the  functions  of  complementary 
angles.  As  a  result  of  the  relations  discussed  in  the  preced- 
ing article,  the  problem  of  computing  the  values  of  the  six 
trigonometric  functions  for  every  angle  between  0°  and  90° 
has  been  reduced  to  that  of  computing  these  values  for  a 
single  one  of  these  functions.  But  we  may  reduce  the  prob- 
lem still  further  by  observing  the  relations  between  the 
functions  of  the  two  acute  angles  of  the  same  right  triangle. 
We  have  (cf.  Fig.  10) 

sm-4  =  -,    cos^  =  -, 
c  c 

cosA  =  -,    sni^  =  -, 
c  c 


COMPLEMENTARY   ANGLES  21 


0  0 

cot  A  =  -,    tan  ^  =  -, 

a  a 

sec^  =  --,    cscij=-, 
h  b 

cscA  =  -,    sec^  =  -, 
a  a 

and  therefore 

^^  ^^       sin  A  =  cos  j5,     tan  ^  =  cot  B,     sec  ^  =  esc  B^ 

cosJ.  =  sin^,     cot  J.  =  tan  ^,     esc  ^  =  sec  ^. 

Since  the  angles  A  and  B  are  complementary,  we  may 
write  these  equations  as  follows : 

sin  (9^°  ~  ^)  =  cos  A,^  cot  (90°  -  ^)  =  tan  A, 
(2)        cos  (90°  -A)  =  sin  Zr^  sec  (90°  -  A)  =  esc  A. 

tan  (90°  -A)  =  cot  A,      esc  (90°  -  ^)  =  sec  ^. 

An  easy  way  to  remember  these  formulae  is  as  follows: 
Let  the  six  functions  be  grouped  into  three  pairs :  sine  and 
cosine,  tangent  and  cotangent,  secant  and  cosecant.  Let  us 
speak  of  either,  function  of  one  of  these  pairs  as  the  cof unction 
of  the  other.  Then,  the  six  formulae  (2)  are  all  included  in 
the  following  statement. 

Any  trigonometric  function  of  the  complement  of  an  angle  A 
is  equal  to  the  cof  unction  of  A. 

It  is  apparent  that  this  theorem  will  enable  us  to  find  the 
trigonometric  functions  of  any  acute  angle  greater  than  45°, 
if  we  know  the  functions  of  all  angles  less  than  45°.  Thus, 
for  instance,  tan  75°  is  equal  to  cot  15°,  sin  82°  is  equal  to 
cos  8°,  etc.  As  a  consequence  of  this  fact  it  is  possible  to 
reduce  the  space  occupied  by  the  tables  of  the  function « 
to  exactly  half  of  what  would  otherwise  be  necessary. 

The  relation  between  the  functions  of  complementary  angles  is  also 
important  in  another  respect.  It  is  this  relation  which  has  given  rise  to 
the  words  cosine,  cotangent,  and  cosecant.  The  cosine  is  the  sine  of  the 
complement.  At  a  time  when  Latin  was  still  the  universal  language  of 
the  scientific  world,  the   cosine  was  therefore  called  complementi  sinus. 


22 


FUNCTIONS  OF  ACUTE  ANGLES 


This  was  later  (in  the  seventeenth  century)  contracted  to  cosinus.     The 
words  cotangent  and  cosecant  originated  in  the  same  manner. 


EXERCISE  VII 

1.  Express  as  functions  of  the  complementary  angles     . 

sin  37°,  cos  62°,  tan  13°,  cot  75,  sec  12°  15',  esc  55°  37'. 

2.  If  the  table  of  values  of  the  functions  is  so  arranged  as  to  give  only 
the  functions  of  angles  less  than  45°,  how  may  we  obtain  the  values  of 

sin  57°,  cos  63°  15',  tan  75^  12',  cot  67°  18'  ? 

3.  What  acute  angle  is  that  whose  sine  is  equal  to  the  sine  of  its 
complement? 

4.  Find  an  acute  angle  for  which  tan^  =  cot  (45°  +  A). 

Hint.    Substitute  for  tan^l  its  equal  cot  (90°  —  A)  and  note  that  two 
acute  angles  with  the  same  cotangent  are  equal  to  each  other. 

5.  Find  an  acute  angle  for  which  sin 2  A  =  cos  (45°  —  A). 

6.  Find  an  acute  angle  for  which  cot  3  A  =  tan  2  A. 

7.  Find  an  acute  angle  for  which  cos^  =  sin  6  ^1. 

8.  Find  an  acute  angle  for  which  sec  2  A  =  esc  7  A. 

11.    The  values  of  the  functions  of  o°,  30°,  45°,  60°,  90°. 

While  we  have  decided  to  postpone  the  general  question  of 
the  arithmetical  calculation  of  the  trigonometric  functions, 
we  have  already  performed  this  calculation  for  a  few  special 
angles,  viz.  :  30°,  45°,  60°  (cf.  Exs.  13,  14  of  Exercise  VI). 
The  figures  there  suggested,  and  the  results  are  as  follows : 


(1) 


^--^ 


(2) 


sin  45°=  cos  45°  =  —^ 

tan  45°=  cot  45°  =  1, 

sec45°=csc45°  =  V2. 

sin  30°  =  cos  60°  =  2-^  =  i, 

c       2 

cos  30°=  sin  60°  =  J  V5, 

tan  30°=  cot  60°  =  -i^  =  -L  =  1  V3, 


cot  30 

sec  30°=  CSC  60° 


1<?V3 
tan  60°  =  V3, 
c 


V3 


2 

/ 

3 

/^ 

\ 

c 

^\ 

/ 

0   \ 

/ 

^    \ 

/ 

II      \ 

M\ 

e  .     \ 

CSC  30°=  sec  60°=  2. 


6=Hc  C 
Fig.  12 


VALUES   FOR   SPECIAL   ANGLES'"  23^ 

An  acute  angle  of  a  right  triangle  can  never  be  0°  or  90°, 
so  that  the  definitions  of  Art.  7  are  not  applicable  to  such  / 
angles.     The  acute  angle  A  may,  however,  approach  either 
0°  or  90°  as  a  limit,  and  if  it  does,  its  functions  in  some  cases 
approach  definite  finite  limits.     By  sin  0°, 
cos  0°,  sin  90°,  cos  90°,  etc.,  we  mean  such 
limits  whenever  they  exist. 

In  Fig.   13,  let  the  angle  A=  FAQ  he 
thought  of  as  decreasing  toward  the  limit  a- 
zero  as  a  result  of  the  rotation  of  the  side 
AQ  around  ^  as  a  center,  while  the  side 
AF  remains  fixed.     Through  any  point  C  of  AF  draw  OR 
perpendicular  to  AF,  and  denote  by  B  the  intersection  of 
on  with  the  rotating  line  AQ. 

As  the  angle  A  approaches  zero,  the  side  a   approaches 
zero  and  c  approaches  b.     Consequently 

8mA=-  approaches  -  or  0,  cos  A  =-  approaches  -  or  1, 

C  0  CO 

tan  A  =  -  approaches  -  or  0,  sec  A  =-■  approaches  -  or  1. 

0  0  0  0 

In  this  sense  we  may  say  that 

sin  0°  -=  0,  cos  0°  =  1,  tan  0°  =  0,  sec  0°  =  1. 

The  function  cot  A  =^     has  no  limit  when  A  approaches  zero, 
a 

For,  the  ratio   -  grows  larger  and  larger  as  A  approaches 
a 

zero,  since  h  remains  fixed  while  a  grows  smaller  and  smaller. 

Clearly,  this  ratio   may  be   made  larger  than   any  number 

however  great,  by  choosing  A  and,  hence,  a  small  enough. 

This  is  expressed  in  symbolic  language  as  follows  : 

cotO°=  00, 

or  in  words  :    The  cotangent  of  an  acute  angle  increases  without 
hound  when  the  angle  approaches  zero  as  a  limit. 

A  similar  argument  holds  for  esc  A=    ,  since  c  approaches 


24  FUNCTIONS  OF  ACUTE  ANGLES 

h  and  a  approaches  zero.     Hence,  in  symbolic  language,  we 

have 

,g.  r  sin  0°  =  0,  tan  0°  =  0,    sec  0°  =  1, 

^    ^  1  cos  0°  =   1,     cot  0°  =  00,     CSC  0°  =  00. 


By  a  similar  argument  the  student  may  deduce  the  follow- 
ing results : 

sin  90°  =  1,    tan  90°  =  oo,    sec  90°  =oo  , 
cos  90°  =  0,   cot  90°  =  0,     CSC  90°  =  1, 


(4)  {-^ 

I  CO 


the  exact  meaning  of  each  of  which  should  be  expressed  in 
words  as  in  the  cases  which  have  just  been  treated  in  extenso. 


CHAPTER   III 

SOLUTION   OF   RIGHT    TRIANGLES  BY   NATURAL 
FUNCTIONS 

12.  Arrangement  and  use  of  the  table  of  natural  functions. 

The  numerical  values  of  the  trigonometric  functions  are 
usually  called  the  natural  functions  to  distinguish  them  from 
the  logarithms  of  these  functions  which  we  shall  study  later. 

Table  *  V  gives  the  numerical  values  of  the  sine,  cosine, 
tangent,  and  cotangent  to  four  decimal  places  for  every  tenth 
of  a  degree  from  0°  to  90°.  The  values  of  the  secant  and 
cosecant  are  omitted  because  they  are  not  used  very  fre- 
quently. They  may  of  course  be  calculated,  whenever 
necessary,  by  the  formulae 

sec  J.  = T,     cscJ.  =  -^ — -.  (Art.  9) 

cos  A  sin  A 

The  following  is  a  sample  portion  of  the  table.  Only  this 
part  of  the  table  will  be  required  for  the  following  illustra- 
tive examples. 


An&le 

N  Stn 

d 

N  Tan 

d 

S  Cot 

d 

NCOS 

d 

35°.0 

0.5736 

14 

0.7002 

26 

1.4281 

52 

0.8192 

11 

55°.0 

.1 

0.5750 

14 

0.7028 

26 

1.4229 

53 

0.8181 

10 

54°.  9 

.2 

0.5764 

15 

0.7054 

m 

1.4176 

52 

0.8171 

10 

.8 

.3 

0.6779 

14 

0.7080 

27 

1.4124 

53 

0.8161 

10 

.7 

.4 

.     .     . 

. 

. 

.6 

.5 

.     .     . 

.     .     . 

.     .     . 

.     .     . 

.5 

N  008 

d 

N  Cot 

a 

N  Tan 

d 

N  Sin 

d 

Angle 

*  See  Logarithmic  and  Trigonometric  Tables,  compiled  by  E.  J.  Wilczynski 
and  H.  E.  Slaught. 

26 


26  SOLUTION   OF  RIGHT  TRIANGLES 

Problem  1.    Find  the  functions  of  35°.2. 

Solution.  In  the  left-hand  column  find  35°. 2.  The  four  numbers 
which  are  printed  in  the  horizontal  row  to  the  right  of  35°.2  are,  /rom  left 
to  right,  the  sine,  tangent,  cotangent,  and  cosine  of  35°.2,  as  indicated  by 
the  name  printed  at  the  head  of  each  of  these  columns.     Therefore 

sin  35°.2  =  0.5764,  tan  35°.2  =  0.7054,  cot  35°.2  =  1.4176, 

cos  35°.2  =  0.8171. 

Problem  2.   Find  the  functions  of  54°.8. 

Solution.  In  the  right-hand  column  find  54°.8.  The  four  numbers 
which  are  printed  in  the  horizontal  row  to  the  left  of  54^.8  are,  from  right 
to  left,  the  sine,  tangent,  cotangent,  and  cosine  of  54°.8,  as  indicated  by 
the  name  printed  at  the  foot  of  each  of  these  columns.     Therefore 

sin  54°.8  =  0.8171,  tan  54°.8  =  1.4176,  cot  54°.8  =  0.7054, 

,    cos  54°.8  =  0.5764. 

Thus  every  number  of  the  table  does  double  duty.  For 
example,  0.5764  is  both  the  sine  of  35°. 2  and  the  cosine  of 
54°.8,  as  it  should  be.     (See  Art.  10,  equations  (2).) 

Angles  less  than  45°  are  given  in  the  left-hand  column  of 
the  table,  and  the  names  of  the  corresponding  functions  are 
found  at  the  top  of  the  page.  Angles  greater  than  45°  are 
given  in  the  right-hand  column  with  the  names  of  the  func- 
tions at  the  bottom  of  the  page. 

The  table  gives  the  values  of  the  functions  only  for  every 
tenth  of  a  degree.  If  the  given  angle  contains  fractional 
parts  of  this  unit,  its  functions  cannot  be  read  directly  from 
the  table.  In  such  cases  we  make  use  of  the  process  of  in- 
terpolation, the  nature  of  which  will  become  apparent  from 
the  following  examples. 

Problem  3.   Find  the  sine  of  35°.17. 

Solution.  This  angle  lies  between  35°.l  and  35°.2.  More  precisely,  it 
lies  -^jj  of  the  way  from  the  former  toward  the  latter.  We  conclude  that 
its  sine  will  be  ^  of  the  way  from 

sin35°.l  =  0.5750  toward  sin  35".2  =  0.5764. 

But  the  difference  d  between  these  last  two  numbers  is  0.0014,  seven 
tenths  of  which  is  equal  to  0.0010  (reduced  to  four  decimal  places). 
Therefore  ^ 

sin  35°.17  =  0  5750  +  0.0010  =  0.5760. 


BY  NATURAL   FUNCTIONS  27 

Problem  4.   Find  the  cotangent  of  35°.l7. 
Solution.     From  the  table  we  find 

cot35°.l=      1.4229 
cot35°.2=      1.4176 
d  =  cot  35°.2  -  cot  35".l  =  -  0.0053  (tabular  difference) 

We  must  add  j\  of  d  to  cot  35M.     Biit^d  =  ~  0.0037. 
Therefore  cot  35M7  =  1.4192. 

We  observe  that  in  problem  3  the  correction  was  positive, 
while  in  problem  4  it  was  negative. 

if  we  always  interpolate  from  the  smaller  toward  the  larger 
angle^  the  correction  will  he  positive  in  the  case  of  sine  and  tan- 
gent^ negative  in  the  case  of  cosine  and  cotangent.  For,  the 
former  two  functions  increase  with  the  angle,  while  the  latter 
two  decrease. 

There  will  never  be  any  serious  danger  of  giving  the  wrong  sign  to 
the  correction,  if  we  cultivate  the  habit  of  running  through  the  numbers 
of  the  table  near  the  place  we  are  using,  so  as  to  see  in  which  direction 
they  are  growing. 

EXERCISE  VIII 

1.  Find  all  of  the  functions  of  15°.3,  28''.7,  63^4,  82°.l. 

2.  Find  sin  37°.24,  cos  62°.19,  tan  53°.42,  cot  27°.13. 

3.  Formulate  in  words  the  principle  upon  which  the  method  of  inter- 
polation is  based.  Is  this  principle  absolutely  exact,  or  is  it  in  the 
nature  of  an  approximation? 

4.  What  are  the  numbers  in  the  four  narrow  columns  of  the  table 
headed  d,  and  what  purpose  do  they  serve  ? 

5.  In  the  arrangement  of  the  table  as  explained,  what  use  has  been 
made  of  the  results  of  Art.  10  ? 

We  have  shown  how  to  find  the  functions  when  the  angle 
is  given.  It  remains  to  show  how  to  find  an  angle  when  one 
of  its  functions  is  known.  The  general  method  will  be  ap- 
parent from  the  following  examples. 

Problem  5.  The  tangent  of  an  unknown  acute  angle  A  is  equal  to 
0.7046.     Find  the  angle  A. 

Solution.  In  the  specimen  table  on  page  25  we  observe  that  the  num- 
ber 0.7046  does  not  occur  in  the  tangent  column.     However,  we  find  there 


28  SOLUTION   OF   RIGHT   TRIANGLES 

the  two  numbers  0.7028  and  0.7054  between  which  0.7046  lies.     Thus  we 

^^^'®  tan  36°.l  =  0.7028, 

tan  A  =  0.7046, 
tan  35°.2  =  0.7054. 

Between  tan  35^.1  and  tan  A,  the  difference  is  0.0018. 

Between  tan  35°.l  and  tan35°.2,  the  difference  is  0.0026. 

Therefore,  tan^  is  ^f  of  the  way  from  tan  35°.l  toward  tan35°.2,  and 
consequently        ^  ^  3^0 -^  +  i|  ^f  qj^^  tenth  of  a  degree, 
or 

A  =  35°.l  +  0°.07  =  35°.17, 
reducing  to  the  nearest  hundredth  of  a  degree. 

Problem  6.   Find  the  acute  angle  A  whose  cosine  is  0.5772. 

Solution.     We  have 

cos  54°.7  =  0.5779, 
cos  A  =  0.5772, 
cos  54°.8  =  0.5764, 
whence 

cos  A  -  cos  54°. 7  =  -  0.0007, 
cos  54^8  -  cos  54°. 7  =  -  0.0015. 

Therefore,  cos  A  is  xV  of  ^he  way  from  cos  54°.7  toward  cos  54°.8.     Hence 

A  =  54°.7  +  jV  of  one  tenth  of  a  degree 
or  A  =  54°.7  +  0°.05  =  54°.75. 

With  a  little  practice,  the  student  will  soon  become  suf- 
ficiently expert  in  the  process  of  interpolation  to  enable  him 
to  perform  this  operation  mentally.  He  should  train  him- 
self with  that  end  in  view. 

If  we  wish  to  find,  by  means  of  our  table,  the  natural 
functions  of  an  angle  which  is  expressed  in  degrees,  minutes, 
and  seconds,  the  minutes  and  seconds  should  first  be  converted 
into  decimal  parts  of  a  degree.  This  may  easily  be  done, 
remembering  that  1'  =  g^^^  of  a  degree  and  1''  =  ^  of  a  minute. 
Table  ^H  ^'^J  ^^  used  to  save  time  in  making  this  trans- 
formation. 

EXERCISE  IX 

1.  Find  the  values  of 

sin  18°  12',  cos  67°  23',  tan  58°  34',  cot  64°  16'.- 

2.  Find  the  acute  angles  for  which 

sin  A  =  0.5673,  tan  C  =  1.7328, 

cos  B  =  0.2791,  cot  D  =  0.8924. 


BY  NATURAL  FUNCTIONS  29 

13.  Solution  of  right  triangles  by  means  of  the  table  of 
natiiral  functions.  A  right  triangle  is  determined  by  any 
two  of  its  parts  (not  counting  the  right  angle)  provided  that 
at  least  one  of  these  parts  is  a  side.  The  relations,  which 
we  have  found  between  the  angles  and  sides,  enable  us  to 
compute  the  remaining  parts  of  a  right  triangle  when  any 
two  such  parts  are  given.  In  order  to  make  sure  of  this 
fact,  let  us  see  what  cases  may  present  themselves. 

If  both  of  the  given  parts  are  sides,  there  are  two  possi- 
bilities. Either  the  two  given  sides  include  the  right  angle 
(Case  1),  or  else  one  of  the  given  sides  is  the  hypotenuse 
(Case  2).  If  one  of  the  given  parts  is  a  side  and  one  is  an 
angle,  we  have  again  two  possibilities,  viz.  :  given  the  hy- 
potenuse and  one  acute  angle  (Case  3) ;  or,  given  one  leg 
and  one  acute  angle  (Case  4).  We  proceed  to  discuss  these 
cases  in  order. 

Case  1.  Given  the  two  sides  of  the  triangle  which  include 
the  right  angle  (the  two  legs^. 

In  our  previous  notation  this  means  that  a  and  h  are  given. 
In  this  case  we  may  first  compute 

tan  A  =  - 

0 

and  then  find  A  from  the  table.     We  may  then  find  c  from 

__     a  h 

sin  A  ~~  cos  A 

in  two  ways  (affording  a  check),  and  B  from 

^  =  90°  -  A. 

Case  2.    Griven  one  leg  and  the  hypotenuse. 

We  may  denote  the  given  leg  by  a..     Then  a  and  c  are. 
given.     The  solution  is  accomplished  by  means  of  the  for- 
mulae 


sin 


A  =  -,  5  =  acot^  =  c?cos^,  i  =  90°-J.. 


30  SOLUTION  OF   RIGHT   TRIANGLES 

Case  3.    G-iven  the  hypotenuse  and  one  acute  angle. 

We  may  denote  the  given  acute  angle  by  A.  Then  A  and 
c  are  given.     The  solution  is  given  by  the  equations 

a  =  e^ix\A^  h  =  c  cos  A,  B  =  90°  —  A, 

Case  4.    G-iven  one  leg  and  one  acute  angle. 

Since  the  knowledge  of  one  acute  angle  implies  that  of  the 
other,  we  may  assume  that  a  and  A  are  the  given  parts.  To 
find  the  remaining  parts,  we  use  the  formulse 

h  =  acoiA,  c  =  — ^  =  _A_^,  ^  =  90°-^. 
sin  A     cos  A 

Our  discussion  has  shown  that  the  methods  at  our  dis- 
posal suffice  to  find  the  remaining  parts  of  a  right  triangle 
when  two  independent  parts  of  the  triangle  (not  counting 
the  right  angle)  are  given.  To  establish  this  fact  was  the 
purpose  of  the  above  classification.  It  is  not  necessary,  nor 
even  desirable,  when  solving  a  numerical  problem  of  this 
sort,  to  find  out  first  under  what  case  it  falls.  In  practice  it 
is  better  not  to  refer  to  this  classification  at  all,  but  to  pick 
out  and  solve  those  among  the  four  equations 

(1)  sin  ^  =  -,  cos  ^  =  -,  tan  A  =  ^,  A  +  B  =  90° 

c  c  b 

which  contain  only  one  unknown  quantity  each.  The  remain- 
ing equations  may  then,  in  most  cases,  serve  as  a  check. 

A  more  complete  check  is  given  by  the  equation 

(2)  a2  +  52^c2. 

In  order  to  avoid  the  inconvenience  of  forming  the  quan- 
tities a\  P^  (^  by  actual  multiplication,  we  have  supplied  a 
table  of  squares  (Table  VI).  The  arrangement  and  use  of 
this  table  will  be  apparent  from  the  following  examples. 

Example  1.     Find  the- squares  of  0.324  and  of  3.24. 

Solution.  In  the  left-hand  column  of  Table  VI  we  find  0.32.  In  the 
same  horizontal  row  with  this  number,  and  in  the  column  headed  4,  we 
find  0.1050.     Therefore 

(0.324)2  ^  0.1050,     (3.24)2  ^  io.50. 


BY  NATURAL  FUNCTIONS  31 

Example  2.     Find  the  squares  of  0.3243,  of  3.243,  and  of  32.43. 

Solution.     From  the  table  we  find 

(0.324)2  ^  0.1050,     (0.325)2  =  0.1056. 

The  difference  between  the  two  squares  is  0.0006.  The  number  0.3243 
is  three  tenths  of  the  way  from  0.324  toward  0.325.  Therefore,  its  square 
will  be  three  tenths  of  the  way  from  0.1050  toward  0.1056.     That  is 

(0.3243)2  3^  0.1050  +  j%  of  0.0006  =  0.1050  +  0.0002  =  0.1052, 
^^^  (3.243)2  =  10.52,     (32.43)2  =  1052. 

Example  3.     Find  the  square  root  of  0.5520. 

Solution.     We  find  from  the  table  that  this  number  is  the  square  of 

0.743.     Therefore  

Vo.5520  =  0.743. 

Example  4.     Find  the  square  root  of  0.5525. 

Solution.     The  table  gives 

Vo:5520  =  0.743,     VO.5535  =  0.744. 

Therefore,  by  interpolation 


Vo:5525  =  0.743  +  j\  of  0.001  =  0.743  +  0.0003  =  0.7433. 

In  engineering  practice,  equation  (2)  is  used  very  exten- 
sively in  connection  with  such  tables  of  squares,  not  merely 
for  checking,  but  for  the  purpose  of  performing  the  original 
calculation.  The  tables  of  Inskip  and  Smoley  are  particu- 
larly convenient  for  this  purpose,  if  the  distances  are  ex- 
pressed in  feet,  inches,  and  thirty-seconds  of  an  inch. 

The  following  examples  illustrate  the  methods  for  solving 
right  triangles. 

Example  5.  In  a  right  triangle,  right  angled  at  C,  given  a  —  3.479, 
b  =  2.321.     Compute  the  remaining  parts  of  the  triangle. 

Solution.     We  find  first 

tan^  =^  =  Mll=lAQS9. 
b      2.321 

The  table  of  tangents  gives 

A  =  56°.2  +  M  of  0°.l  =  56°.30, 

reducing  to  the  nearest  hundredth  of  a  degree,  as  usual.     The  tables  of 
sines  and  cosines  give 

sin  A  =  0.8320,     cos  A  =  0.5548. 


32  SOLUTION  OF  RIGHT  TRIANGLES 

We  compute  next 

«         ^'^^^    =4.181. 


sin  A      0.8320 

and  use  the  equation  c  cos  ^  =  6  as  a  check.     We  find 

c  cos  ^  =  2.320,     b  =  2.321. 

The  two  members  of  the  check  equation  do  not  agree  exactly,  but  we 
have  no  right  to  expect  absolute  agreement.  All  of  the  numbers  used  in 
the  calculation  are  merely  approximations,  giving  us  the  values  of  the 
functions  to  the  nearest  unit  of  the  fourth  decimal  place.  In  combining 
several  such  approximate  numbers,  the  error  may  occasionally  exceed 
two  or  three  units  of  the  last  decimal  place. 
Finally  we  find 

B  =  QO°  -A  =  33°.70. 

We  may  exhibit  this  solution  more  compactly  as  follows.  The  figures 
in  parentheses  indicate  the  order  in  which  the  various  results  are  ob- 
tained. 

Formulce.  tan  ^  =  -,  c  =  — ^,  B  =  90°  -  A. 

b  sin  A 

Check.  c  cos  A  =  b. 

Given  '  "  ^  ^'^^^     ^^^    ®^^  ^  "  ^'^^^^  ^^^  ^  "  ^'^^^   ^'^^ 


b  =  2.321     (2)    cos  A  =  0.5546  (6)    c  cos  ^  =  2.320    (8)  Check, 
tan  A  =  1.4989  (3)  A  =  56°.30  (4)  B  =  33°.70  (9). 

.  Example  2.  In  a  right  triangle,  right  angled  at  C,  given  c  =  5.783, 
A  =  42°.39.     Compute  the  remaining  parts  of  the  triangle. 

Solution.     Formulae,     a  =  c  sin  vl,     b  =  c  cos  A,     B  =  90°  —  A. 
Check.  a2  +  62  =  c\ 

Given     '  '  =  ^-^^^     <^1)  «  =  ^'^^^    (^> 

\A  =42°.39  (2)  6  =  4.271    (7) 

B  =  47°.61  (8)  a2  ^  15.20  (8) 

sin  A  =  0.6742  (4)  b^  =  18.24  (9) 

cos  A  =  0.7386  (5)  a^  +  b^  =  33.44  (10)  |  q^^^^ 

c2  =  33.44  (11)  J 

Remark.  The  quantities  a^  b^,  c^  required  for  the  check  were  obtained 
from  the  table  of  squares.  When  no  such  table  is  available,  it  is  usually 
desirable  to  write  the  check  equation  in  the  form 

a2  =  c2-62  =  (c-  b)(c-\-b), 

since  this  form  of  the  equation  reduces  by  one  the  number  of  multiplica- 
tions required. 


BY   NATURAL   FUNCTIONS  33 

In  this  example,  the  check  computation  would  then  yield 
a2=  15.202,  c-b  =  1.512, 

c  +  b=  10.054, 
c^-b^=  15.202     Check. 

EXERCISE  X 

In  each  of  the  following  right  triangles,  right  angled  at  C,  two  parts 
are  given.  Compute  the  remaining  parts  and  check.  Also  check  by 
means  of  a  graphic  solution  to  provide  against  gross  errors. 

^"1.  a  =  27,       A  =  25°.l.  /  5.  c=  604.5,  A  =  47°.53. 

i/2.  a  =  34.5,    e  =  52.8.  6.  a  =  8.695,  b  =  7.321. 

3.  a  =  2.781,  b  =  3.056.  7.  b  =  62.78,  c  =  81.39. 

4.  &  =  87.95,  ^  =  55°.36.  8.  jB=29°.58,  c  =  2354. 

9.   A  gravel  roof  slopes  three  fourths  of  an  inch  per  horizontal  foot. 
What  angle  does  it  make  with  a  horizontal  plane  ? 

10.  The  pitch  of  a  gable  roof  is  the  quotient  obtained  by  dividing  the 
height  of  the  ridge-pole  above  the  garret  floor  by  the  width  of  the  garret. 
What  is  the  pitch  of  a  gable  roof  covering  a  garret  38  feet  wide,  if  the 
ridge-pole  is  15  feet  above  the  garret  floor,  and  what  angle  does  the  roof 
make  with  a  horizontal  plane  ? 

11.  At  a  time  when  the  sun  was  55°  above  the  horizon,  the  shadow  of 
a  certain  building  was  found  to  be  112  feet  long.  How  high  is  the 
building  ? 

12.  The  side  of  a  regular  decagon  is  3.471  feet.  Find  the  radii  of  the 
inscribed  and  circumscribed  circles. 

13.  The  side  of  a  regular  polygon  of  n  sides  is  equal  to  a.  Find 
formulae  for  the  radii  of  the  inscribed  and  circumscribed  circles. 


CHAPTER  IV 

DISCUSSION    OF   SOME    DEVICES    FOR    REDUCING    THE 
LABOR   INVOLVED  IN   NUMERICAL    COMPUTATIONS 

14.  The  number  of  decimal  places.  As  we  attempted  to 
point  out  in  Chapter  I,  every  number  obtained  as  a  result  of 
measurement  is  really  an  approximation.  If  we  measure  the 
distance  between  two  dots  on  our  drawing  board,  by  means 
of  a  carefully  constructed  scale  which  reads  to  the  fiftieth 
part  of  an  inch,  we  may  still  estimate  half  of  ^his  smallest 
scale  unit  with  the  naked  eye.  Let  us  assume  that  the 
divisions  of  the  scale  are  reliable,  that  the  ruler  is  very 
nearly  straight,  that  the  dots  are  very  small,  and  that  we  are 
using  the  greatest  of  care  in  our  measurement.  We  may 
then  concede  that  the  result  of  such  a  measurement  (say  5.34 
inches)  is  accurate  to  the  nearest  y^  of  an  inch.  This 
means  that  no  number,  with  two  figures  to  the  right  of  the 
decimal  point,  is  as  close  to  the  true  value  as  5.34.  It  means 
that  5.33  is  certainly  too  small  and  that  5.35  is  certainly  too 
large.  It  does  not  mean  that  the  true  value  is  exactly  5.34 
inches,  but  that  the  true  value  lies  between  5.335  and  5.345 
inches. 

When  we  record  the  result  of  such  a  measurement,  the 
number  of  decimal  places  which  we  write  (two  in  this 
example)  is  an  indication  of  the  degree  of  precision 
which  we  claim  for  the  result.  In  this  connection,  let  us 
note  emphatically  that  a  zero,  when  obtained  as  the  last  digit 
of  the  measure  of  a  quantity,  should  never  be  suppressed. 
Suppose,  for  instance,  that  in  the  above  example  we  had 
obtained  5.30  inches  as  the  result  of  our  measurement. 
This  means  that  we  are  certain  that  the  true  value  of  the 
distance   lies   somewhere   between  5.295  and  5.305  inches. 

34 


ACCURACY   OF   RESULTS  35 

If  we  were  to  record  this  result  as  5.3  inches,  suppressing 
the  final  zero,  we  should  be  giving  the  erroneous  impression 
that  we  had  measured  the  distance  only  to  the  nearest  tenth 
of  an  inch  and  that  it  might  have  any  value  between  5.25 
and  5.35  inches. 

Thus,  the  number  of  decimal  places,  which  we  use  in  record- 
ing the  result  of  a  measurement^  is  an  indication  of  the  degree 
of  precision  which  we  attribute  to  this  result. 

This  being  so,  we  are  guilty  of  negligence  whenever  we 
express  such  a  result  by  a  decimal  with  fewer  places  than  we 
are  able  to  guarantee.  For  we  are  thus  throwing  away 
knowledge  which  we  have  actually  had  in  our  possession. 
But  it  would  be  dishonest  to  express  our  result  with  more 
decimal  places  than  we  can  guarantee.  For  we  should  then 
be  tending  to  mislead  others  into  thinking  our  measurements 
more  accurate  than  they  really  are. 

We  may  estimate  the  degree  of  precision  of  a  number,  as  we 
have  just  done,  on  an  absolute  scale.  But  clearly  it  will 
usually  be  more  reasonable  to  adopt  as  a  measure  of  pre- 
cision the  ratio  of  the  ^'  probable  error  "  of  the  measurement 
to  the  total  magnitude  of  the  quantity  measured.  If  we  do 
this,  an  error  of  one  foot  in  a  thousand  is  to  be  regarded 
as  of  no  greater  importance  than  an  error  of  ywq'Q  ^^  ^^ 
inch  in  an  inch.  In  either  case  we  may  say  that  it  is  an 
error  of  -^^  of  one  per  cent. 

Whenever  we  properly  record  the  result  of  a  measurement 
by  a  number  consisting  of  four  digits,  no  matter  where  the 
decimal  point  may  be  placed,  this  means,  in  the  light  of  our 
preceding  discussion,  that  the  error  of  the  last  digit  is  guar- 
anteed to  be  less  than  half  a  unit  of  the  last  decimal  place. 
Now,  the  smallest  number  expressed  by  four  digits  is  1000. 
Let  us  suppose  that  the  exact  value  of  our  unknown  quan- 
tity is  w  =  1000  units,  but  that  as  a  result  of  our  measure- 
ment we  have  found  a;  =  999.5  units.  Then  our  error  is 
2  oVo'  ^^  A"  ^^  ^^^  P®^  cent,  of  the  total  magnitude  measured. 
The  largest  number  expressible  by  four  digits  is  9999.     If 


36  NUMERICAL  COMPUTATIONS 

the  exact  value  of  x  is  9999  and  our  error  of  measurement  is 
half  a  unit,  the  error  will  be  to  the  total  magnitude  measured 
as  \  is  to  9999,  or  as  1  is  to  19998.  It  will  be  an  error  of 
about  2^0  ^^  ^^®  P®^  ^®^^  ^^  ^^  whole.  Thus,  four  digits 
are  certainly  sufficient  to  express  the  result  of  a  measurement 
if  its  degree  of  accuracy  does  not  exceed  ^^  of  one  per  cent. 
Now,  most  of  the  ordinary  operations  of  surveying  fall  well 
within  this  limit,  so  that  four  decimal  places  are  usually 
sufficient  to  express  the  results  obtained  by  the  surveyor. 

15.  The  accuracy  of  a  sum,  difference,  product,  or  quotient 
of  two  numbers  obtained  by  measurement.  '  It  is  clear  that 
the  sum  or  difference  of  two  numbers  can  have  no  greater 
precision  than  the  less  accurate  of  the  two  numbers.  Con- 
sequently, it  is  useless  and  misleading  to  retain  more  decimal 
places  in  one  term  of  a  sum  or  difference  than  in  the  others. 

In  forming  a  product  we  are  apt  to  do  a  great  deal  of  use- 
less work  if  we  fail  to  remember  that  the  factors,  and  there- 
fore the  product,  are  mere  approximations.  Suppose  we 
have  measured  the  sides  a  and  5  of  a  rectangular  field  to  the 
nearest  hundredth  of  a  foot  and  have  found  a  =35.67  ft., 
5  =  86.72  ft.  To  find  the  area  of  the  field  we' form  the 
Oft  79  product  ah.  The  ordinary  method  (shown  in 
„_'   _     the  margin)  gives  the  result  3093.3024  square 

'  feet.     But  if  we  allow  the  result  to  stand  in 

-n/^or.       thls  form,  we  shall  exhibit  either  our  ignorance 

O2032  A       '  ^  ^  ^    ^  '  •  17 

. „^  or  our  desire  to  create  a  false  impression,     l"  or 

9fini  R  *^^^  would  seem  to  indicate  a  result  precise  to 

^OQ^  ^r>9J.     ^^  nearest  .^^^^^^  of  a  square  foot,  whereas  it  is 

uncertain  by  more  than  a  whole  square  foot  as 

we  shall  now  show. 

In  fact,  the  equations  a  =35.67  ft.,  5  =  86.72  ft.  merely 
mean  that  a  and  h  are  between  the  limits 

35.665  <a<  35.675,   86.715  <h  <  86.725 
respectively,  so  that  the  area  must  be  between  the  limits 

86.715  X  Zb.m^  <ah<  86.725  x  35.675 
or  3092.685475  <  ah  <  3093.919375. 


LABOR-SAVING  DEVICES  37 

The  uncertainty  in  the  value  of  ah  is  therefore  so  great 
that  no  digit  to  the  right  of  the  decimal  point  has  any  real 
significance.  Even  the  last  digit  preceding  the  decimal 
point  is  not  certain,  so  that  we  are  even  slightly  overstating 
our  accuracy,  if  we  write  simply 

ah  =  3093  square  feet. 

Thus,  we  see  that  the  product  of  two  approximate  four- 
place  numbers  is  to  be  regarded  as  accurate  to  no  more  than 
four  places.  Consequently,  it  is  a  waste  of  time  and  labor  to 
actually  work  out  all  of  the  partial  products  in  the  multipli- 
cation.    We  might  abbreviate  the  process  as  follows ; 


86.72 

or  better,  by 

86.72 

35.67 

writing  the 

35.67 

,6. 

more  important 

2602. 

52. 

partial  products 

434. 

434. 

first; 

52. 

2602. 

6. 

3094.  3094. 

The  fact  that  we  find  3094,  instead  of  3093,  need  not  dis- 
turb us,  since  we  have  seen  from  our  above  considerations 
that  the  last  digit  is  actually  uncertain  to  the  extent  of  one 
unit. 

This  process,  which  is  sometimes  called  ahhreviated  multi- 
plication^  is  obviously  preferable  to  the  ordinary  method, 
since  it  does  away  with  the  labor  of  first  finding  numbers 
which  must  afterwards  be  discarded. 

Similar  remarks  may  be  made  for  division.  More  generally, 
whenever  we  are  dealing  with  numbers  whose  first  four  digits 
only  can  be  regarded  as  certain,  it  is  wise  to  abbreviate  all 
of  our  calculations  correspondingly.  It  should  be  remarked, 
however,  that,  in  certain  exceptional  cases,  very  exact  results 
may  be  obtained  from  inaccurate  data  and  vice  versa.  But 
this  is  not  the  place  for  a  discussion  of  such  cases. 

16.  Labor-saving  devices.  We  have  seen  that  extensive 
calculations,  even  with  four-place  numbers,  are  apt  to    be 


38  NUMERICAL  COMPUTATIONS 

troublesome  and  laborious.  In  many  problems  it  is  neces- 
sary to  use  five-  six-  or  seven-place  numbers.  In  such  cases, 
the  amount  of  labor  required  becomes  excessive,  even  if  we 
make  use  of  the  abbreviated  method  of  multiplication  and 
division. 

A  very  important  aid  in  performing  such  calculations  is 
furnished  by  certain  tables,  such  as  tables  of  squares  (of 
which  Table  VI  is  an  example),  tables  of  cubes,  and  recip- 
rocals of  numbers,  etc.  Crelle's  Tables,  which  are  merely 
a  systematically  arranged  and  extensive  set  of  multiplication 
tables,  are  particularly  valuable. 

A  second  great  aid  to  numerical  calculation  is  furnished 
by  calculating  machines,  which  are  now  being  used  very 
extensively  in  commercial  as  well  as  scientific  work.  The 
ordinary  cash  register  is  one  of  the  simplest  of  these 
machines. 

Graphic  methods  for  solving  numerical  problems  consti- 
tute a  third  great  class  of  aids  to  calculation.  These  meth- 
ods have,  in  recent  times,  been  modified  and  extended,  so 
as  to  become  capable  of  greater  accuracy,  and  for  many 
problems  no  other  method  of  solution  is  known.  The 
slide  rule,  which  may  be  classed  either  with  the  graphic 
methods  or  among  the  calculating  machines,  has  become 
so  popular  among  engineers  as  to  exclude,  in  their  work, 
almost  all  other  methods  of  calculation.  (See  Arts.  29 
and  30.) 

But  the  most  important  of  all  of  these  labor-saving  de- 
vices, without  which  the  slide  rule  and  many  similar  con- 
trivances cannot  even  be  thoroughly  understood,  is  the 
method  of  calculation  by  logarithms. 

17.  Definition  of  logarithms.  It  is  apparent,  from  what 
has  been  said,  that  the  real  cause  of  the  laboriousness  of 
extensive  calculations  lies  in  the  operations  of  multiplication 
and  division.  Addition  and  subtraction,  even  of  numbers 
with  many  places,  are  comparatively  easy.  It  was  this  fact 
which  caused  John  Napier  (1550-1617)  and  Jobst  Burgi 


DEFINITION  OF  LOGARITHMS  39 

(1552-1632)*  to  consider  the  possibility  of  devising  a 
method,  by  means  of  which  addition  and  subtraction  might 
be  made  to  do  the  work  of  multiplication  and  division.  The 
method  which  they  invented  for  this  purpose  is  essentially 
equivalent  to  the  one  which  we  shall  now  explain,  though 
very  different  from  it  in  form.  We  must  remember  that  the 
notations  of  modern  algebra,  to  which  we  are  accustomed 
and  which  are  of  the  greatest  assistance  to  us  in  our  mathe- 
matical arguments,  are  the  results  of  a  century-long  process 
of  development.  This  process  was  far  from  complete  in 
Napier's  and  Biirgi's  time.  The  greatness  of  the  achieve- 
ment of  these  men  can  only  be  properly  appreciated  when 
judged  from  the  standpoint  of  the  mathematical  knowledge 
of  those  days.f 

From  our  present  point  of  view  the  possibility  of  reduc- 
ing the  operations  of  multiplication  to  that  of  addition  is 
an  immediate  consequence  of  a  familiar  fact  of  algebra. 
This  fact,  embodied  in  the  formula 

states  that  the  product  of  two  powers  of  the  same  base  is 
itself  a  power  of  that  base,  whose  exponent  is  equal  to  the 
sum  of  the  exponents  of  the  two  original  powers.  We  may 
express  this  fact  by  the  statement  that  to  the  multiplication 
of  two  powers  of  the  same  base  corresponds  the  addition  of 
their  exponents.  By  this  simple  remark,  multiplication  is 
actually  converted  into  addition  for  all  such  numbers  as  are 
known  powers  of  a  common  base.  '^^ 

We  shall  assume  that  the  fixed  number  a,  the  base  of  our 
system,  is  positive  and  different  from  unity.  If  the  expo- 
nent X  is  a.  positive  integer,  there  will  be  comparatively  few 

*  Napier  was  of  Scotch  and  Burgi  of  Swiss  nationality.  Biirgi's  discovery 
of  logarithms  was  unquestionably  independent  of  Napier's  and  was  made  at 
about  the  same  time.  But  Napier's  book  "  Mirifici  Logarithmorum  canonis 
descriptio,"  containing  an  account  of  his  method  was  published  in  1614,  six  years 
earlier  than  Biirgi's  "  Arithmetische  und  Geometrische  Progress-Tabidn." 

t  For  an  excellent  account  of  the  history  of  logarithms  see  Cajori  in  The 
American  Mathematical  Monthly,  Vol.  20  (1913). 


40  NUMERICAL  COMPUTATIONS 

numbers  which  can  be  regarded  as  powers  of  a.  Suppose, 
for  example,  that  a  =  2.  Then  2,  4,  8,  16,  32,  etc.,  are 
integral  powers  of  2,  but  1,  3,  5,  6,  7,  9,  10,  etc.,  are  not. 
But,  from  our  previous  study  of  algebra,  we  are  acquainted 
with  the  fact  that  the  symbol  a^  may  be  defined,  not  merely 
for  the  case  when  the  exponent  is  a  positive  integer,  but  also 
when  the  exponent  is  any  positive  or  negative  rational  num- 

ber  of   the  form   ±  -    (j9  and  q  being  integers),    or   zero. 

These  definitions  are  as  follows : 

If  a;  is  a  positive  integer  (x  =  p)^ 

I.  a'  =  a'P  ==  a  •  a  •  a  '••  (a  product  of  p  factors  each  equal 
to  a). 

If  X  is  a  positive  rational  fraction  f  re  =  -  j, 

II.  a-  =  a^/«  =  Va^  =  Q-Vay, 
where  Va  means  the  positive  qih.  root  of  a. 

If  ic  is  a  negative  rational  fraction  {x  = J, 

III.  a-  =  a-^/«  =  -^=— . 

Finally,  if  rr  =  0, 

lY.  a-=aO  =  l. 

Now,  there  is  nothing  remarkable  about  the  fact  that  we 
have  been  able  to  define  the  symbol  a^  in  all  of  these  cases. 
We  might  have  done  that  in  many  different  ways.  But  it 
i8  remarkable  that,  if  we  adopt  these  particular  definitions, 
the  formulae 

(1)  o" '  a^  =  a^+^ 

(2)  ^=^a^-y, 
^  ^  ay 

(3)  (a^)^  =  «'^ 

turn  out  to  be  true,  not  only  when  the  exponents  are  posi- 
tive integers  but  in  all  of  the  four  cases  for  which  we  have 


DEFINITION   OF   LOGARITHMS  41 

defined  the  symbol  a^.  That  this  should  be  so,  is  of  course 
not  merely  a  fortunate  coincidence.  It  is  due  to  the  fact 
that  the  definitions  II,  III,  IV  were  deliberately  chosen  in 
such  a  way  as  to  insure  the  universal  validity  of  formulae 
(1),  (2),  and  (3).  These  formulae,  which  are  collectively 
known  as  the  three  index  laws^  are  fundamental  for  the  fol- 
lowing discussion. 

Equations  I,  II,  III,  IV  suffice  to  define  the  symbol  a* 
whenever  a;  is  a  rational  number.  Now,  every  irrational 
number  can  be  approximated,  as  closely  as  we  may  desire,  by 
means  of  a  decimal  fraction  ;  and  this  decimal  fraction 
(which  is  a  rational  number)  will  take  the  place  of  the  original 
irrational  number  in  all  numerical  calculations.  We  may 
define  a%  when  x  is  irrational,  as  follows  : 

Let  a^i  be  the  closest  approximation,  to  the  irrational  num- 
ber a;,  which  is  possible  by  means  of  a  decimal  fraction  with 
only  one  figure  to  the  right  of  the  decimal  point.  Let  x^  be 
the  closest  approximation  possible  by  means  of  a  decimal 
fraction  with  only  two  figures  to  the  right  of  the  decimal 
point.  Let  a:g,  ic^,  •••  a;„,  •••  be  similar  approximations  with 
3,  4,  '"  n  figures  after  the  decimal  point.  The  sequence  of 
rational  numbers        ^    ^    ^    ..  „    ... 

has  the  irrational  number  a;  as  a  limit.  Then  a*  is  defined 
to  be  the  limit  of  the  second  sequence  of  numbers 

a%  a%  a^3,  •••  «%  •••  .* 

As  an  example,  consider  a  =  10,  a;  =  V2.     We  have 

x^  =  1.4,  x^  =  1.41,  x^  =  1.414,  x^  =  1.4142,  etc., 

ax,  =  101-4,  a^.  =  101-41,  a-3  =10i-*iS  a^^  =  lO^-^i^^  etc. 

By  10^^  WQ  mean  the  limit  approached  by  the  numbers  of 
this  second  sequence. 

For  practical  purposes,  however,  V2  is  replaced  by  one  of 
the  approximations  1.4,  1.41,  1.414,  etc.,  namely,  the  first 
one  which  is  sufficiently  accurate  for  the  particular  problem 

*  These  limits  exist,  but  it  would  carry  us  too  far  to  prove  this  fact. 


42  NUMERICAL  COMPUTATIONS 

under  consideration  ;  and  10^2  jg  replaced  by  the  first  one 
of  the  numbers  lO^-*,  IQi**!,  etc.,  which  is  sufficiently  close  to 
the  true  value  for  the  purposes  of  the  problem  in  question. 

It  may  be  shown  that,  if  the  above  definition  of  a^  for 
irrational  values  of  x  be  adopted,  the  index  laws  will  hold  also 
for  irrational  exponents. 

We  may  now  state,  without  formal  proof,  a  theorem  which 
is  fundamental  in  the  theory  of  logarithms,  in  so  far  as  the 
very  existence  of  logarithms  depends  upon  it.  This  theorem 
is  as  follows  : 

If  a  is  a  positive  number  different  from  unity^  there  exists 
one  and  only  one  exponent  x  {positive,  negative,  or  zero^,  such 
that  ^^^^ 

where  N  is  any  positive  number. 

Although  we  state  this  theorem  without  proof,  the  student 
may  easily  convince  himself  of  its  great  plausibility  by  a 
process  which,  if  carried  out  to  its  logical  conclusion,  would 
constitute  a  proof.  Suppose,  for  instance,  that  a  =  2  and 
that  iV=  1000.     We  have 

26  =  32,  26  =  64,  27  =  128,  2^  =  256,  2^  =  512,  2io  =  1024. 

We  conclude  that  the  value  of  x  for  which 

2'  = 1000 

must  be  between  9  and  10.  Now  2i/2  =  V2  =  1.4142  .... 
Therefore 

29-5  =  29.21/2=  512  X  1.4142  =  724.1. 
But  210  =  1024, 

so  that  X  must  lie  between  9.5  and  10.  We  may  obtain 
closer  and  closer  limits  between  which  x  must  lie  by  con- 
tinuing this  process,  and  thus  ultimately  show  that  there 
exists  a  number  x  (as  a  limit  of  a  sequence),  for  which 
2'  —  1000.  This  argument  at  the  same  time  indicates  a 
process  by  means  of  which  the  exponent  x  may  be  calculated 
to  any  desired  number  of  decimal  places. 


I 


PROPERTIES  OF  LOGARITHMS  43 

We  are  now  ready  to  define  a  logarithm. 

The  logarithm  of  any  positive  number  N,  with  respect  to  the 
base  a,  is  the  exponent  of  the  power  to  which  the  base  a  must  he 
raised  in  order  to  obtain  the  number  N.* 

In  other  words^if  a^  =  iV, 

we  say  that  x  (the  exponent)  is  the  logarithm  of  iV  with 
respect  to  the  base  a.  In  symbols  we  write  this  same  state- 
ment as  follows  :   '  ^  _  i^^.   ^ 

Example.     Since  5^  =  125,  we  have  log^  125  =  3. 

EXERCISE   XI 

1.  What  are  the  logarithms  of  2,  4,  8,  16,  32,  64,  128  with  respect  to 
the  base  2?    Write  out  each  of  these  results  in  symbols ;  thus,  logg  4  =  2. 

2.  What  are  the  logarithms  of  3,  9,  27,  81,  243  with  respect  to  the 
base  3? 

3.  What  are  the  logarithms  of  10,  100,  1000,  10,000  with  respect  to 
the  base  10  ? 

4.  What  are  the  logarithms  of  3,  9,  27,  81,  243  with  respect  to  the 
base  27? 

5.  What  are  the  logarithms  of  1,  I,  \,  ^^,  -^j,  ^\-g  with  respect  to  the 
base  3  ? 

6.  What  are  the  values  of  2*,  3*,  4',  10*  when  x  is  equal  to  zero? 
What,  then,  is  the  logarithm  of  1  with  respect  to  each  of  the  bases  2,  3, 
4,  10? 

7.  What  is  the  logarithm  of  1  with  respect  to  any  base  a? 

8.  What  is  the  logarithm  of  any  number  with  respect  to  itself  as 
base? 

9.  Find,  approximately  to  two  decimal  places,  the  number  whose 
logarithm,  with  respect  to  the  base  2,  is  equal  to  1.5. 

18.  The  properties  of  logarithms.  Those  properties  of 
logarithms  which  are  of  importance  for  the  purposes  of  nu- 
merical calculation,  are  immediate  consequences  of  the 
index  laws  and  of  the  definition  of  logarithms.     In  fact,  we 

*The  word  logarithm  is  derived  from  the  Greek  Ao-yos  or  logos,  meaning  pro- 
portion or  ratio,  and  apie^o?  or  arithmos,  meaning  number.  The  reason  for 
choosing  this  name  will  be  apparent  from  the  theorem  stated  in  Exercise  XII, 
Ex.  4. 


44  NUMERICAL   COMPUTATIONS 

may  write  each  of  two  positive  numbers,  M  and  iV,  in  the 
form 

(1)  M^a"",     N=^ay, 

so  that,  in  accordance  with  the  definition  of  logarithms, 

(2)  x  =  \og,M,     y=\og,N. 

According  to  the  first  index  law  (Art.  17,  equation  (1)), 
the  product  of  M  and  N  is  equal  to 

whence,  by  the  definition  of  logarithms, 

log„  (TJfiV^)  =  x  +  y  =  log,  M-\-  log„  K 

The  theorem  expressed  by  this  formula  may  obviously  be 
extended  to  any  number  of  factors.     Hence, 

I.    The   logarithm  of  a  product  is  equal  to  the  sum  of  the 
logarithms  of  its  factors. 

From  (1)  we  obtain  by  division, 

making  use  of  the  second  index  law  (Art.  17,  equation  (2)). 
Therefore,  by  the  definition  of  logarithms, 


*       loga \^—\  =  x-y  =  log,  M-  log, iV, 

a  result  which  may  be  formulated  as  follows  : 

II.  The  logarithm  of  a  quotient  is  equal  to  the  logarithm 
of  the  dividend  minus  the  logarithm  of  the  divisor. 

The  same  fact  may,  of  course,  be  stated  in  the  equivalent 
form:  the  logarithm  of  a  fraction  is  equal  to  the  logarithm  of  the 
numerator  minus  the  logarithm  of  the  denominator. 

According  to  the  third  index  law  (Art.  17,  equation  (3)), 
we  have  (^xy^^^x^ 


PROPERTIES   OF   LOGARITHMS  45 

Therefore,  we  find  from  (1) 

•    M^  =  aP\ 

or,  by  the  definition  of  logarithms, 

log^  Mp  =px=p  log^M. 
Consequently, 

III.  The  logarithm  of  the  p^^  power  of  a  number  M  is  ob- 
tained by  multiplying  the  logarithm  of  M  by  p. 

Since  the  third  index  law  is  true  whether  p  be  an  integer 
or  a  fraction,  the  last  theorem  has  the  following  corollary, 

obtained  by  putting  p  equal  to  — : 

IV.  The  logarithm  of  the  n*^  root  of  a  number  Mis  obtained 
by  dividing  the  logarithm  of  M  by  n. 

The  content  of  the  two  equations 

a^  =  a,     a^  =  1 

may  be  stated  as  follows  : 

V.  The  logarithm  of  any  number^  with  respect  to  itself  as 
base,  is  equal  to  unity. 

VI.  77ie  logarithm  of  unity,  with  respect  to  any'  base,  is 
equal  to  zero. 

It  is  clear  now  how  logarithms  will  serve  to  reduce  the 
operations  of  multiplication  and  division  to  those  of  addition 
and  subtraction.  Suppose  that  we  have  at  our  disposal  a 
table  of  logarithms.  To  multiply  iKf  by  iV  we  look  up  the 
logarithms  of  these  numbers  from  the  table  and  add  them 
together.  We  then  find  the  number  whose  logarithm  is 
equal  to  this  sum  by  again  referring  to  the  table  ;  this  num- 
ber is  the  product  MN.  To  divide  M  hj  JSf  we  proceed  in 
the  same  way,  except  that  in  this  case  we  form  the  difference 
log  M—  log  iV  instead  of  the  sum. 


46  NUMERICAL   COMPUTATIONS   . 

EXERCISE  XII 

1.  lUog,,  2  =  0.30103,  logio  3  =  0.47712,  find  log,,  12,  logio  (I),  logio  e^) , 
log  10  y/Q. 

2.  Express,  in  terms  of  \ogaP  and  log^  g,  the  following  quantities  : 


loga   (pY),    loga    (^),    l0g« 


3.  Prove  the  truth  of  the  following  statement.  If  logio  x  is  expressed 
as  a  decimal  fraction  (x  being  a  positive  number  greater  than  unity), 
tlie  logarithm  of  10*  a:  (k  being  a  positive  integer)  will  differ  from  logio  a; 
only  in  its  integral  part. 

4.  Prove  the  theorem  :  If  the  numbers  a,,  a^  Og,  ...  are  in  geo- 
metrical progression,  their  logarithms  are  in  arithmetical  progression. 

5.  Prove  the  equation 

iog„  x  +  Vx^-1  ^  2  log^  (x  +  v^s^n:). 


CHAPTER  V 

CALCULATION  BY  LOGARITHMS 

19.  Common  logarithms.  With  scarcely  an  exception,  the 
civilized  nations  of  all  times  have  made  use  of  the  decimal 
system  for  expressing  numbers,  both  in  the  spoken  and  in  the 
written  language.*  For  this  reason,  the  number  10  is  es- 
pecially well  adapted  to  serve  as  base  for  a  system  of 
logarithms.  Logarithms  with  respect  to  the  base  10  are 
usually  known  as  common  logarithms ;  they  are  also  some- 
times called  Briggsian  logarithms,  in  honor  of  Henry  Briggs 
(1556-1630),  f  who  constructed  the  first  table  of  common 
logarithms. 

In  this  book  we  shall  have  very  little  occasion,  hereafter,  to 
speak  of  any  except  the  common  logarithms.  We  shall 
therefore  agree  to  abbreviate  the  symbol  logjQ  iVto  log  iV, 
the  base  10  being  understood  when  no  other  base  is  men- 
tioned explicitly. 

The  positive  integral  powers  of  10,  such  as  10,  100,  1000, 
etc.,  the  negative  integral  powers  of  10,  such  as  0.1,  0.01, 
0.001,  etc.,  and  the  zero  power  of  10,  which  is  equal  to  1, 
are  the  only  numbers  whose  common  logarithms  are  integers. 
The  logarithms  of  all  other  numbers  have  an  integral  and  a 
fractional  part. 

The  fractional  part  of  the  logarithm  is  called  the  mantissa, 
while  the  integral  part  of  the  logarithm  is  known  as  its  char- 
acteristic. 

*Itis  usually  admitted  that  the  predominance  of  the  decimal  system  over  all 
others  is  due  to  the  fact  that  the  normal  human  heing  has  ten  fingers.  This 
opinion  has  certainly  been  generally  held  since  the  time  of  Aristotle, 

t  Briggs  was  the  first  Savilian  Professor  of  Geometry  at  Oxford.  According 
to  Ball  (see  Ball's  Primer  of  the  History  of  Mathematics) ,  Briggs  was  also  the 
first  to  make  systematic  use  of  the  decimal  notation  in  working  with  fractions. 

47 


48  CALCULATION   BY   LOGARITHMS 

20.  Properties  of  the  mantissa.  We  consider  the  mantissa 
and  the  characteristic  separately  because,  in  practice,  the 
method  for  finding  the  characteristic  of  a  logarithm  is  entirely 
different  from  that  employed  for  finding  its  mantissa.  The 
reason  for  this  will  appear  from  the  following  discussion. 

Let  us  consider  an  example.  From  the  definition  of  a 
common  logarithm,  we  know  that 

(1)  log  </10  =  log  10^  =  1  log  10  =  f  1  =  0. 25000. 

Now  it  is  not  difficult  to  compute  VlO  by  elementary 
methods.  We  may,  for  instance,  first  compute  VlO  to  as 
many  decimal  places  as  we  desire,  and  then  extract  the  square 
root  of  the  result.  We  find,  in  this  way,  to  five  decimal 
places, 

(2)  ^10  =  1.77828 
or,  if  we  combine  (1)  and  (2), 

(3)  log  1.77828  i=  0.25000. 

From  the  theorem  about  the  logarithm  of  a  product,  we 
conclude 
log    17.7828  =  log     (1.77828  x  10)  =  log  1.77828  +  log    10 

=  0.25000  +  1  =  1.25000, 
log    177.828  =  log  (1.77828  x  100)  =  log  1.77828  +  log  100 

=  0.25000  +  2  =  2.25000, 

We  observe  that  the  numbers  1.77828,  17.7828,  177.828, 
etc.,  contain  the  same  succession  of  digits,  and  differ  from 
each  other  only  in  the  position  of  the  decimal  point.  Their 
logarithms,  on  the  other  hand,  whose  values  we  have  just 
calculated,  differ  from  each  other  only  in  the  value  of  the 
characteristic. 

Again,  if  we  make  use  of  the  theorem  about  the  logarithm 
of  a  quotient,  we  find  from  (3) 

log  0.177828  =  log  lill^  =  0.25000  -  1, 

log  0.0177828  =  log  ^4^^^  =  0.25000  -  2, 
8  ^100 


DETERMINATION   OF  THE   CHARACTERISTIC         49 

Now  the  negative  quantities,  which  appear  on  the  right 
members  of  these  equations,  are  not  written  in  the  form 
which  we  ordinarily  use  for  negative  quantities.  Thus,  for 
instance,  we  have  found  the  value  of  log  0.0177828  to  be 
0.25000  —  2,  a  result  which  we  should  ordinarily  write  in  the 
form  —  1.75000,  to  which  it  is  obviously  equal.  If  we  agree 
to  write  every,  negative  logarithm  in  this  unusual  form, 
as  a  difference  between  a  positive  proper  fraction  and  an 
integer,  thus  making  its  fractional  part  positive,  we  gain  the 
advantage  that  the  mantissas  will  be  the  same  for  any  two 
numbers  which  contain  the  same  succession  of  digits,  even  if 
none  of  these  digits  appear  to  the  left  of  the  decimal  point. 
We  avoid,  in  this  way,  the  necessity  of  using  two  different 
tables  of  mantissas,  one  for  numbers  greater,  and  one  for 
numbers  less,  than  unity. 

Let  us'  recapitulate  the  result  of  our  discussion  in  two 
formal  statements. 

I.  We  agree  to  express  the  logarithm  of  any  positive  number 
Nin  such  a  form  that  its  mantissa  shall  he  positive. 

This  can  be  done  whether  log  N  is  positive  or  negative, 
that  is,  whether  N  be  greater  or  less  than  unity.  In  the 
latter  case,  the  negativeness  of  log  N  is  brought  about  en- 
tirely by  means  of  the  negative  characteristic. 

As  a  consequence  of  this  agreement,  the  following  state- 
ment will  be  true  in  all  cases. 

II.  If  two  numbers  contain  the  same  succession  of  digits^  that 
is,  if  they  differ  only  in  the  position  of  the  decimal  point,  their 
logarithms  will  have  the  same  mantissa  and  will  differ  only  in 
the  value  of  the  characteristic. 

It  is  for  this  reason  that  the  tables  give  only  the  mantissas 
of  the  logarithms  and  that,  in  looking  up  the  mantissas,  we 
pay  no  attention  to  the  position  of  the  decimal  point  in  the 
given  number. 

21.  Determination  of  the  characteristic.  The  characteristic 
of  a  logarithm  is  easily  determined  by  inspection.     Its  value 


50  CALCULATION   BY   LOGARITHMS 

depends  merely  on  the  position  of  the  decimal  point.     Since 
we  have 

100  ^  1^  101  =  10,  102  ^  100,  103  =  1000,  etc., 
or 

log  1  =  0,  log  10  =  1,  log  100  =  2,  log  1000  =  3,  etc., 

we  draw  the  following  conclusions : 

If  1<  iVr<  10,  then  0  <  log  iV^<  1.  .• .  log  iV^  has  the 
characteristic  0. 

If  10  <  iV^<  100,  then  1<  log  iV^  <  2.  .  • .  log  iV^  has  the 
characteristic  1. 

If  100  <  N<  1000,  then  2  <  log  iV^<  3.  .-.  log  iV^  has 
the  characteristic  2. 

'    IflO*<iV^<10*+i,thenA;<logiV^<A;  +  l.     .-.logiVhas 
the  characteristic  k. 

We  may  formulate  these  results  as  follows: 

III.  If  k  18  a  positive  integer,  and  if  the  number  iV  lies  be- 
tween 10*  and  10*  """i,  the  characteristic  of  log  N  is  equal  to  k. 

Since  such  a  number  N  has  k  -\-\  digits  to  the  left  of  the 
decimal  point,  we  obtain  the  following  rule : 

IV.  If  N  is  any  number  greater  than  1,  the  characteristic  of 
its  logarithm  is  one  less  than  the  number  of  digits  in  its  integral 
part. 

The  student  is  advised  to  make  but  little  use  of  this  rule 
on  account  of  its  mechanical  character.  Statement  III  pro- 
vides a  better  method  (less  mechanical  and  easier  to  re- 
member) for  determining  the  characteristic. 

It  remains  to  show  how  to  find  the  characteristic  of  log  N 
when  iV<  1. 

If  .1  <  iV'<  1,  then  -  1<  log  iV^<  0.  .  • .  log  iV^  has  the 
characteristic  —  1. 

If  .01  <  N<  .1,  then  -  2  <  log  iV<  -  1.  .-.  log  N  has 
the  characteristic  —  2. 


USE  OF  THE  TABLE  OF  LOGARITHMS       51 

If  .001<iV'<.01,then-3<logiV^<  -  2.     .-.  log  iV^ has 
the  characteristic  —  3. 


li^^<J^<^,,then-ik+l)<\ogir<-k.     .-.log 

iV  has  the  characteristic  —  (A:  +  1). 

Examination  of  this  table  leads  to  the  following  two  state- 
ments, either  of  which  may  be  used  to  determine  the  char- 
acteristic of  log  JV  when  iV<  1. 

If^  k  is  a  positive  integer,  and  if  the  number  N  lies  between 

——  and  — - — ,  the  characteristic  of  log  N  is  —  Ck  -f-  1). 
10*  10*  "''1 

If  N  is  less  than  1,  and  is  expressed  as  a  decimal  fraction 
having  k  zeros  betyjeen  the  decimal  point  and  the  first  significant 
figure,  then  the  characteristic  of  the  logarithm  of  Nis  —  (k  -\-  1). 
In  one  of  our  illustrations  we  had  found 

log  0.0177828  =  0.25000-2: 
We  must  never  write  this  in  the  form  "^ 

log  0.0,1,77828  =  -  2.25000,     ^    ?  '  ' 
since  only  the  characteristic  "is  negative  and  not  the  fractional  part. 
Some  computers  use  the  notation 

log  0.0177828  =  2.25000 ; 

but  for  most  purposes  it  is  preferable  to  write  "" 

log  0.0177828  =  8.25000  -  10, 
and  similarly 

log  0.177828  =  9.25000  -  10. 

In  other*  words,  in  actual  practice,  we  write  a  positive  char- 
acteristic 10  —  Z:  in  place  of  the  negative  characteristic  —  k,  and 
theyi  subtract  10  from  the  whole  logarithm. 

22.  Arrangement  and  use  of  the  table  of  logarithms.     We 

have  already  mentioned  the  fact  that  the  table  of  logarithms 
gives  only  the  mantissas.  The  characteristics  must  be  sup- 
plied by  the  computer  by  the  methods  of  Art  21. 

The  table  which  we  shall  ordinarily  use  (Table  I)  gives 
the  mantissa,  for  every  number  from  1  to  9999,  to  five  decimal 
places. 


62 


CALCULATION   BY   LOGARITHMS 


In  order  to  explain  the  arrangement  of  this  table,  we  shall 
reprint  a  small  portion  of  it,  and  solve  a  number  of  typical 
examples,  chosen  in  such  a  way  as  to  require  only  this  part 
of  the  table  for  their  solution. 


N 

0 

1 

2 

3 

4 

6 

6 

7 

8 

9 

PP 

34242 

.  439 

635 

830 

262 
459 
655 
850 

282 
479 
674 
869 

301 
498 
694 
889 

321 
518 
713 
908 

341 
537 
733 
928 

361 
557 
753 
947 

380 

577 
772 
967 

400 
596 
792 
986 

420 

616 

811 

*005 

19 

20 

220 
221 
222 
223 

1 
2 
3 
4 
5 
6 
7 
8 
9 

L9 

3.8 

5.7 

7.6 

9.5 

11.4 

13.3 

15.2 

17.1 

2.0 

4.0 

6.0 

8.0 

10.0- 

12.0 

14.0 

16.0 

18.0 

Problem  1.     Find  the  logarithm  of  221.4. 

Solution.  To  find  the  mantissa  we  ignore  the  decimal  point.  We 
read  down  the  left-hand  column  of  the  table  (headed  N)  until  we  find 
the  first  three  digits  of  our  number,  viz.,  221.  The  numbers  printed  in 
the  same  horizontal  row  with  221  are,  in  order,  the  mantissas  of  the 

logarithms  of  2210,    2211,    2212,   ,  2219,   as    indicated   by  the 

number  at  the  head  of  each  of  the  next  ten  columns.  To  save  space, 
however,  the  first  two  digits  of  the  mantissa  are  never  printed  more  than 
once  in  each  row.  In  our  case  we  find  the  mantissa,  from  the  column 
headed  4,  to  be  .34518.  Since  221.4  is  between  100  =  10^  and  1000  =108, 
the  characteristic  is  2.     Therefore 

log  221.4  =  2.34518. 

Problem  2.    J'ind  log  22.39. 

Solution.  Looking  for  the  mantissa  as  before,  we  find  *005.  The 
asterisk  indicates  that  the  first  two  digits  of  the  mantissa  are  not  34,  as 
one  might  suppose,  but  35.  The  reason  for  this  appears  clearly  from  the 
table.     Therefore 

log  22.39  =  1.35005. 

If  the  number  N  contains  more  than  four  digits,  its  loga- 
rithm cannot  be  read  directly  from  the  table.  But  it  may 
be  found  by  interpolation.  We  illustrate  this  process  by.  an 
example. 


USE  OF  THE  TABLE  OF  LOGARITHMS      53 

Problem  3.     Find  log  222.73. 

Solution.  From  the  table  we  find,  supplying  the  characteristics  our- 
selves, 

log  222.70  =  2.34772 
log  222.80  =  2.34792 
Tabular  difference  =  0.00020  =  20  units  of  the  fifth  decimal  place. 

Since  222.73  is  ^  of  the  way  from  222.70  toward  222.80,  we  add  ^^  of 
the  tabular  difference  to  log  222.70.     Therefore 

log  222.73  =  2.34772  +  ^  of  0.00020, 
or 

log  222.73  =  2.34772  +  0.00006  =  2.34778. 

The  auxiliary  tables  in  the  margin,  headed  P  P  (abbre- 
viation for  proportional  parts),  facilitate  the  process  of  in- 
terpolation. 

Thus,  in  problem  3,  we  refer  to  the  auxiliary  table  with  20  (the 
tabular  difference)  at  its  head.     In  the  third  row  we  find  ^^^  of  20  or  6,0. 

It  remains  to  show  how  to  find  the  number  when  its  loga- 
rithm is  given. 

Problem  4.  Given  log -^  =  9.34857  -  10.  Find  the  value  of  N  to 
five  significant  figures. 

Solution.  The  characteristic  of  log  N  is  9  —  10  or  —  1.  Therefore, 
the  number  N  must  be  between  10-^  =  0.1  and  10"  =  1.  Consequently, 
the  decimal  point  will  precede  the  first  significant  figure  of  N. 

The  mantissa  34857  does  not  occur  in  the  table,  but  it  falls  between 
the  two  tabular  mantissas  34850  and  34869. 
Thus  we  have : 

9.34850  -  10  =  log 0.22310  (from  the  table), 
9.34857  -  10  =  logN, 

9.34869  -  10  =  log  0.22320  (from  the  table), 
so  that  N  lies  between  0.22310  and  0.22320. 

We  observe  that  log  N  lies  jV  of  the  way  from  log  0.22310  toward 
ogO.22320.  Therefore,  N lies  rV  of  the  way  from  0.22310  toward  0.22320. 
That  is, 

N=  0.22310  +  xV  of  10  units  of  the  fifth  decimal  place. 
But 

^-^  of  10  units  =  {%  units  =  3|f  units  =  4  units. 

Therefore 

N^  0.22310  +  0.00004  =  0.22314. 


54  CALCULATION  BY  LOGARITHMS 

Also  in  this  inverse  problem  (to  find  the  number  when  its 
logarithm  is  given)  interpolation  is  aided  by  the  auxiliary 
tables  in  the  margin. 

Thus,  in  problem  4,  the  tabular  difference  is  19.  The  difference  be- 
tween log  N  and  the  smaller  of  the  two  tabular  logarithms,  between 
which  log  N  lies,  is  7.  The  auxiliary  table  with  19  at  its  head,  shows 
that,  among  the  tenths  of  19,  that  one,  which  comes  closest  to  the  value 
7,  is  the  fourth.  Consequently,  N  is  j\  of  the  way  from  0.22310  toward 
0.22320.  Therefore,  up  to  five  decimal  places,  iV=  0.22310  +  0.00004 
=  0.22314. 

23.  Cologarithms.  Since  we  obtain  the  same  result  whether 
we  divide  iV  by  iHf,  or  multiply  iV  by  — ,  we  may,  in  a  loga- 
rithmic calculation,  add  the  logarithm  of  —  insteiad  of  subtract- 
ing log  M.  The  logarithm  of  ~-  is  called  the  cologarithm  of  M. 
Therefore 

colog  M=  log  -—  =  log  1  —  log  M=  —  log  M, 

since  log  1  is  equal  to  zero. 

Cologarithms,  like  logarithms,  are  written  with  positive 
mantissas.  Consequently,  the  cologarithm  of  a  number  is 
most  easily  found  by  subtracting  its  logarithm  from  zero, 
written  in  the  form  10.00000  —  10,  as  in  the  foUowiLg 
example. 

Problem  5.    Find  the  cologarithm  of  222.73. 

Solution. 

10.00000  -  10 
log222.73  =  2.34778 


colog  222.73  =  7.65222  -  10 
It  is  easy  to  perform  this  operation  of  subtraction  from 
10.00000  —  10  mentally.  There  is  no  gain,  however,  from  the 
use  of  cologarithms  when  we  are  dealing  with  a  quotient  of 
only  two  numbers.  A  real  advantage  is  gained  by  the  in- 
troduction of  cologarithms,  when  more  than  two  logarithms 
are  to  be  combined  by  addition  and  subtraction.  For  the 
logarithms  which  are  to  be  subtracted  we  then  substitute 


EXTRACTION  OF   ROOTS  65 

cologarithms,  enabling  us  to  complete  the  operation  by  a 
single  addition. 

It  often  happens,  just  as  in  the  case  of  forming  a  cologa- 
rithm,  that  we  wish  to  subtract  a  logarithm  from  another 
smaller  one.  In  all  such  cases  we  change  the  form  of  the 
minuend  by  adding  and  subtracting  10,  or  some  convenient 
multiple  of  10,  as  in  the  following  example. 

32  34 

Problem  6.     Compute  -—^. 

472.3 

Solution.     We  find  from  Table  I, 

log  32.34  =  1.50974, 
log  472.3  =  2.67422. 

In  order  to  subtract  the  latter  logarithm  from  the  former,  we  write 
log  32.34  =11.50974 -10,* 
log472.3  =  2.67422 

log?Mi  =  8.83552 -10 
^472.3 

Hence,  from  the  table,       ?^^  =  0.068473. 
'  472.3 

24.    Extraction  of  roots  by  means  of  logarithms.     Since 
log  V^  =  log  x'^P  =  -  log  X, 

it  is  easy  to  extract  roots  of  any  order  by  means  of  logarithms. 
If  the  characteristic  of  logo;  is  not  negative,  no  further  re- 
mark is  necessary.  If  log  x  is  negative,  we  proceed  as  in  the 
following  example : 


Problem  7.     Compute  by  logarithms :  V.53760,  ^.53760,  and  V.53760. 
Solution.     Iog0.53760  =  9.73046-10. 

log  V.58760  =  I  log  0.53760  =  i  (19.73046  -  20)  =  9.86523  -  10. 
log  \/.53760  =  i  log  0.53760  =  i  (29.73046  -  30)  =  9.91015  -  10. 
log  V:53760  =  ^  log  0.53760  =  i  (49.73046  -  50)  =  9.94609  -  10. 
Therefore,  from  Table  I, 

V.53760  =  .73322,  V.53760  =  .81312,  V.53760  =  .88326. 


*  A  computer  with  some  experience  will  refrain  from  actually  writing  the 
logarithm  in  the  form  11.50974  —  10.  It  is  easy  for  him  to  carry  out  the  calcula- 
tion as  though  it  were  so  written. 


56  CALCULATION  BY  LOGARITHMS 

25.    Logarithmic  calculations  which  involve  negative  numbers. 

We  have  only  defined  the  logarithms  of  positive  numbers. 
But  this  suffices  for  our  purposes.  Clearly,  vrhen  we  com- 
pute a  product  or  quotient,  its  numerical  value  may  be  found 
first,  without  paying  any  attention  to  the  signs  of  the  various 
factors.  Afterwards,  the  proper  sign  (  +  or  —  )  may  be  pre- 
fixed to  the  result  according  as  there  is  an  even  or  an  odd 
number  of  negative  factors. 

The  easiest  way  to  keep  a  count  of  the  negative  factors  is 
to  use  the  method,  introduced  by  Gauss,*  of  writing  the 
letter  n  immediately  after  a  logarithm  which  corresponds 
to  a  negative  number.  In  forming  a  sum  or  difference  of 
logarithms,  we  write  an  n  after  the  result  only  if  an  odd 
number  of  the  separate  logarithms  is  affected  by  an  n. 

Example.     If  iV  =  -  222.73,  we  write 

log  iV=  2.34778  n. 

EXERCISE  XIII 

1.  Making  use  of  the  tables,  find  log  3726,  log  67.43,  log  729800, 
log  0.3896,  log  0.008527. 

2.  Making  use  of  the  tables,  find  log  32653,  log  76.431,  log  879450, 
log  0.045723,  log  0.0059426. 

3.  By  means  of  the  tables,  find  the  numbers  whose  logarithms  are 
3.84522,  1.68079,  8.89064  -  10,  7.12548  -  10,  2.27068. 

4.  By  means  of  the  tables,  find  the  numbers  whose  logarithms  are 
3.89067,  9.24110-10,  1.52195. 

5.  Given  a  =  3.1572,  b  =  7.2916,  c  =  45.731.  Compute  by  logarithms 
the  values  of  ab,  be,  ca. 

6.  With  the  same  values  of  a,  b,  c  compute  —  • 

c 


7.   With  the  same  values  of  a,  b,  c  compute  V^  • 


8.  Compute  the  volume  of  a  hemispherical  dome  if  its  diameter  is 
150.32  feet.     (Volume  of  a  sphere  of  radius  r  is  f  xr^.) 

*C.  F.  Gauss  (1777-1855)  was  without  question  one  of  the  greatest  and  taoat 
versatile  mathematicians  of  all  times.  He  was  director  of  the  observatory  and 
professor  of  astronomy  at  Gottingen  from  1807  to  the  end  of  his  life. 


LOGARITHMS   OF   THE   TRIGONOMETRIC   FUNCTIONS     57 

9.  If  a  sum  of  money  P  (the  principal)  is  earning  interest  at  the 
rate  of  r  %  a  year,  and  if  the  interest  is  added  to  the  principal  at  the  end 
of  each  year,  show  that  the  amount,  at  the  end  of  n  years,  will  be 

In  this  case  the  interest  is  said  to  be  compounded  annually. 

10.  Find  the  amount  on  $157.38  for  7  years  at  ^\%  compound 
interest. 

11.  How  much  money  must  I  put  into  the  bank  at  3  %  compound 
interest,  so  that  the  amount  may  be  $  500  at  the  end  of  five  years  ? 

26.  The  logarithms  of  the  trigonometric  functions.  In  solv- 
ing right  triangles  by  means  of  logarithms,  we  frequently 
have  to  find  the  logarithm  of  a  trigonometric  function  of  an 
angle.  It  would  be  very  burdensome  if  we  had  to  look  up 
first,  in  the  table  of  natural  functions,  the  value  of  the  func- 
tion and  then,  from  the  table  of  mantissas,  find  its  loga- 
rithm. In  order  to  avoid  this  complication,  there  has  been 
constructed  an  additional  table  which  enables  us  to  find 
directly  the  values  of  the  logarithms  of  the  sine,  cosine,  tan- 
gent, and  cotangent  for  every  angle  between  0°  and  90°. 
These  quantities  are  denoted  by  the  symbols  log  sin  or 
L.  Sin,  log  cos  or  L.  Cos,  etc.,  and  are  pronounced  log  sine, 
log  cosine,  log  tangent,  and  log  cotangent. 

In  the  tables  which  accompany  this  book.  Table  II. gives 
the  values  of  the  logarithms  of  the  trigonometric  functions 
directly,  to  five  decimal  places,  for  every  minute  of  arc.  If 
the  angle  contains  fractional  parts  of  a  minute,  we  obtain  its 
functions  from  the  table  by  interpolation. 

The  arrangement  of  this  table  resembles  that  of  the  table 
of  natural  functions  so  closely,  that  it  is  unnecessary  to  de- 
scribe it  in  detail.  It  should  be  noted,  however,  that  in 
this  table  the  characteristics  of  the  logarithms  are  also  given. 
But  since  the  natural  sines  and  cosines  of  all  acute  angles, 
and  the  tangents  of  all  angles  less  than  45°,  are  proper  frac- 
tions, these  characteristics  are  negative  and  have  been  ex- 
pressed in  the  form  9  —  10,  8  —  10,  etc.  The  continually  re- 
curring — 10  has  not  been  printed^  and  should  be  supplied  by 


58 


CALCULATION  BY  LOGARITHMS 


the  computer.  It  is  understood,  once  for  all,  that  10  is  to  be 
subtracted  from  all  of  the  logarithms  in  the  first,  second,  and 
fourth  columns  of  the  table,  while  the  logarithms  printed  in 
the  third  column  are  provided  with  their  correct  character- 
istics and  require  no  such  modification. 

The  process  of  interpolation  may  be  applied  to  the  table  of 
logarithms  of  the  trigonometric  functions  in  the  same  way 
as  to  the  table  of  natural  functions  or  to  the  table  of  loga- 
rithms of  numbers. 

The  following  examples  are  intended  to  illustrate  the  ap- 
plication of  Table  II. 

Example  1.     Find  log  sin,  log  cos,  log  tan,  log  cot  of  41°  15'  35". 
Solution.     For  convenience  in  interpolation  convert  35"  into  decimal 
parts  of  a  minute.     Then  41"  15'  35"  =  41°  15'.58. 

We  find,  from  the  table,  the  following  material: 

41° 


/ 

LSiN 

» 

LTan 

C  D 

LCoT 

LCos 

D 

PP 

15 
16 

9.81911 
9.81926 

16 

9.94299 
9.94324 

25 

0.05701 
0.05676 

9.87613 
9.87601 

12 

45 

44 





~1 

2 
3 
4 

5 
6 

7 
8 
9 

15 

1.5 

3.0 

4.5 

6.0 

7,5 

9.0 

10.6 

12.0 

13.5 

LCo8 

D 

LCoT 

0  D 

LTan 

LSiN 

D 

1 

48° 

We  conclude : 

log  sin  41°  15'.58  =  9.81911  +  .58  of  15  units  of  the  5th  decimal  place, 
log  tan  41°  15'.58  =  9.94299 +  .58  of  25  units  of  the  5th  decimal  place, 
log  cot  41°  15'.58  =  0.05701  -  .58  of  25  units  of  the  5th  decimal  place, 
log  cos41°15'.58  =  9.87613  -  .58  of  12  units  of  the  5th  decimal  place. 

We  may  use  the  marginal  tables  of  proportional  parts  to  complete  the 
interpolation.     Thus,  the  table  headed  15,  shows  that  .5  of  15  is  7.5  and 


THE   ACCURACY   OF   FIVE-PLACE   TABLES  59 

.08  of  15  is  1.2,  and  consequently  .58  of  15  is  8.7  or  9  units  of  the  fifth 
decimal  place.     Therefore 

log  sin  41°  15'.58  =  9.81920  -  10. 
In  the  same  way  we  find 

log  tan  41°  15'.58  =  9.94314  -  10,  log  cot  41°  15'.58  =  0.05686, 
log  cos  41°  15'.58  =  9.87606  -  10. 

Example  2.     Find  the  logarithms  of  the  functions  of  48°  44'.42. 

Solution.  This  angle  is  the  complement  of  that  of  Example  1.  Hence 
each  of  its  functions  is  equal  to  the  corresponding  cofunction  of  41°  15'.58, 
and  the  values  obtained  are  the  same  as  in  Example  1  with  the  name  of 
each  function  changed  to  the  corresponding  cofunction. 

Just  as  in  the  table  of  natural  functions,  these  values,  for  angles 
greater  than  45°,  may  be  obtained  directly  from  the  table  by  reading  the 
degrees  of  the  angle  at  the  foot  of  the  page,  the  minutes  in  the  right- 
hand  column,  and  the  name  of  the  function  at  the  foot  of  each  of  the 
four  columns.     We  find,  in  this  way, 

log  sin  48°44'.42  =  9.87606  -  10,   log  cot  48°44'.42  =  9.94314  -  10, 
log  tan  48°  44'.42  =  0.05686,  log  cos  48°  44'.42  =  9.81919  -  10. 

Example  3.     Given  log  tan  A  =  0.53219.     Find  A. 

Solution.  The  given  logarithm  does  not  appear  anywhere  in  the  col- 
umn at  the  foot  of  which  is  printed  the  name  L  Tan.  But  we  do  find 
in  this  column  j^^  ^^^^  730  33,  ^  ^  53312, 

log  tan  73°  39^  =0.53259. 
Tabular  difference  for  1'  =  0.00047. 
The  given  value  of  log  tan  A  is  -^j,  or  y^^,  of  the  way  from  the  first 
toward  the  second  of  these  tabular  logarithms.     Therefore 

A  =  73°  38M5 

27.  The  accuracy  of  five-place  tables.  As  we  have  said 
repeatedly,  the  number  of  decimal  places  used  in  stating  the 
result  of  a  measurement  is  to  be  regarded  as  an  indication  of 
its  degree  of  precision.  We  shall  ordinarily  wish  to  make 
all  calculations,  based  upon  such  measurements,  with  a  suffi- 
cient number  of  decimal  places  to  avoid  introducing  inaccu-. 
racies  which  might  have  an  appreciable  influence  upon  the 
results,  i.e.  an  influence  comparable  with  that  produced  by 
the  unavoidable  errors  of  observation. 

Five-place  tables  are  quite  accurate  enough  to  satisfy  this 
condition  in  almost  all  problems  of  engineering  and  natural 


60  CALCULATION  BY  LOGARITHMS 

science.  In  fact,  in  most  problems  of  this  kind,  the  distances 
are  not  measured  so  accurately  as  to  exclude  an  error  of  one 
twentieth  of  one  per  cent  of  their  value,  and  the  angles  are 
read  to  the  nearest  minute  only.  But  five-place  tables  are 
far  more  accurate  than  this.  In  fact,  a  distance  expressed 
by  a  five-place  number  presupposes  an  accuracy  of  at  least 
^J-g-  %,  and  an  angle  may  usually  be  determined  from  five- 
place  logarithms  of  its  functions  with  an  error  of  not  more 
than  2  or  3  seconds  of  arc. 

We  must  not  forget,  however,  that  the  degree  of  accuracy 
of  a  table  is  not  the  same  in  all  parts  of  the  table,  and  that 
we  must  use  our  judgment  in  the  selection  of  the  formulae 
which  we  wish  to  use  in  solving  a  problem. 

28.  The  trigonometric  functions  of  angles  near  o°  or  90°. 

Thus,  for  instance,  if  we  wish  to  determine  an  angle  for  which 
log  cos^  =  9.99998  —  10,  our  table  cannot  furnish  an  accu- 
rate result.  We  find,  by  referring  to  the  table,  that  A  may 
have  any  value  between  0°  29'  and  0°  36'. 

A  small  angle  cannot  be  determined,  with  any  degree  of 
accuracy,  from  the  value  of  its  cosine. 

In  the  same  way,  we  see  that  an  angle  very  close  to  90° 
cannot  he  determined  accurately  from  the  value  of  its  sine. 

In  most  cases  we  shall  be  able  to  modify  the  formula, 
which  we  are  using,  in  such  a  way  as  to  avoid  this  difficulty. 
If,  for  instance,  the  angle  A  (known  to  be  very  small)  is  to 
be  determined  from  the  value  of  its  cosine,  we  shall  seek 
some  other  formula  as  a  solution  of  the  same  problem  by 
means  of  which  the  angle  A  can  be  determined  from  the 
value  of  its  sine  or  tangent.  The  problem  then  reduces  to 
that  of  finding  a  small  angle  when  its  sine  or  tangent  is 
given.  If  we  again  refer  to  our  table,  we  find  that  this 
problem  also  gives  rise  to  a  difficulty.  The  method  of  inter- 
polation, which  we  ordinarily  use,  becomes  both  cumbersome 
and  inexact  in  the  case  of  such  small  angles,  because  the 
tabular  differences  are  very  large  and  change  very  rapidly 
from  one  place  in  the  table  to  another. 


THE  LOGARITHMIC   OR  GUNTER  SCALE  61 

In  order  to  meet  this  difficulty,  we  have  provided  a  sepa- 
rate table  (Table  III),  giving  the  values  of  the  logarithmic 
functions  for  every  second  of  arc  from  0°  0'  to  0°  3'  and  from 
89°  57'  to  90°,  and  for  every  ten  seconds  from  0°  to  2°  and 
from  88°  to  90°. 

Another  method  of  meeting  this  difficulty,  preferable  in 
some  respects,  will  be  explained  in  the  second  part  of  this 
book  (Art.  85);  it  involves  the  auxiliaries  S  and  2^  (Table 
IV). 

EXERCISE  XIV 

1.  Find  log  sin  15^  6',  log  cos  35°  13',  log  tan  57°  28',  log  cot  76"  44'. 

2.  Find  log  sin  18°  23'.35,  log  tan  41°  4'^6'.27,  log  cos  64°  17'  43",  log  cot 

25°  12'  38". 

3.  Find  the  angles  for  which 

log  sin  A  =  9.42553  -  10,        log  cos  B  =  9.60618  -  10 
log  cot  C  =  9.68497  -  10,        log  tan  D  =  0.14193. 

/  4.   Find  the  angles  for  which 

log  cot  A  =  0.11157,  log  tan  B  =  9.75465  -  10, 

log  cos  C  =  9.68334  -  10,         log  sin  D  =  9.56652  -  10. 

5.  Find  log  sin  0°  2'  15",  log  tan  1°  10'  22". 

6.  Find  the  angles  for  which 

log  sin  A  =  5.83170  -  10,        log  tan  B  =  8.32313  -  10. 

7.  Find,  by  logarithms,  the  angle  A,  if  tan  A  =  a/b  and  a  =  1.2291, 
b  =  14.950. 

29.  The  logarithmic  or  Gunter  scale.  Ko  graphical  process 
is  more  familiar  than  the  addition  and  subtraction  of  line- 
segments,  and  this  process  may  evidently  be  used  as  a  sub- 
stitute for  addition  and  subtraction  of  numbers.  Since  addi- 
tion of  logarithms  corresponds  to  multiplication  of  numbers, 
we  may  find  the  logarithm  of  a  product  graphically  by  add- 
ing line-segments,  whose  lengths  are  equal  to  the  logarithms 
of  the  factors. 

In  order  to  do  this,  we  must  have  some  means  for  actually 
finding  a  line-seguient  whose  length  shall  be  equal  to  the 
logarithm  of  a  given  number. 


62  CALCULATION  BY   LOGARITHMS 

Let  us  take  a  line-segment  of  convenient  length,  say  10 
centimeters,  as  unit  of  length.  In  terms  of  this  unit,  the 
whole  distance  (10  centimeters  =  100  millimeters)  repre- 
sents log  10,  since  the  logarithm  of  10  is  equal  to  unity. 
If  we  count  all  distances  from  the  left-hand  end  of  the  line, 
we  may  label  the  right-hand  end  10  to  indicate  that  this 
distance  represents  log  10.  The  left-hand  end  will  then  be 
labeled  1,  because  log  1=0. 

From  the  table  of  logarithms  we  have,  to  two  decimal  places, 

logl=0.00,      log2=0.30,    log3  =  0.48,    log4  =  0.60, 

(1)  log5  =  0.70,      log6  =  0.78,    log7  =  0.85,    log8  =  0.90, 

log  9  =  0,95,  log  10  =  1.00. 
We  mark  the  points  on  our  line-segment  whose  distances 
from  the  left-hand  end,  measured  in  terms  of  the  whole  line 
as  unit,  are  in  order  equal  to  log  2,  log  3,  log 4,  ...  log  9, 
and  label  them  2,  3,  4,  ...  9,  respectively.  If  the  whole 
line-segment  is  10  centimeters  long,  these  points  will,  on 
account  of  (1),  be  at  distances  30,  48,  60,  70,  78,  85,  90,  95 
millimeters,  respectively,  from  the  left-hand  end  of  the  line- 
segment  (cf.  Fig.  14). 

1 2 3  4  5       6      7     8    9   10 

Fio.  14 

A  scale  constructed  in  this  way  is  called  a  logarithmic 
scale,  and  its  usefulness  for  purposes  of  calculation  was  first 
pointed  out  by  Edmund  Gunter*  in  1620.  It  enables  us 
to  find  a  line-segment  equal  in  length  to  the  logarithm  of 
any  number  between  1  and  10.  It  is  easy  to  see  how,  by 
means  of  such  a  scale  and  a  pair  of  dividers,  multiplication 
and  division  may  be  reduced  to  the  simple  graphical 
processes  of  adding  and  subtracting  line-segments. 

30.  The  slide  rule.  Some  years  before  1630,  William 
OuGHTRED  f  noticed  that  the  use  of  the  dividers  might  be 
avoided  by  constructing  two  equal  logarithmic  scales  (Scales 

♦Professor  of  astronomy  in  Greshara  College,  London  (1581-1626). 
t  OuQHTRED  (1575-1600)  was  a  fellow  of  King's  College,  Cambridge. 


THE   SLIDE   RULE 


63 


A  and  B  of  Fig.  15),  capable  of  sliding  by  each  other,  as 
indicated  in  the  figure.* 


i 


5       6 


t^ 


9  10 


"I 1 — 1 1 — T~ 

6      7    8    9    10 


Fig.  15 

The  use  of  this  simple  bit  of  apparatus  for  the  purpose  of 
multiplication  and  division  will  be  apparent  from  the  follow- 
ing examples  : 

To  multiply  2  by  3.  Place  scale  B  in  such  a  way  that  its  left-hand 
index  (i.e.  the  division  marked  1)  falls  directly  under  the  division 
marked  2  on  scale  A.  Directly  above  the  division  marked  3  on  scale  B, 
we  shall  find,  on  scale  A,  the  product  which  (of  course)  is  6.  To  justify 
this  process  it  suffices  to  note  that  it  is  equivalent  to  adding  the  log- 
arithm of  3  to  that  of  2. 

Fig.  15  shows  scales  A  and  B  in  the  proper  position  for  the  purposes 
of  this  example. 

To  divide  6  hy  3.  Under  the  division  6  of  scale  A,  place  division  3  of 
scale  B.  Over  the  division  1  of  scaled  we  shall  find  the  quotient  (f  =  2) 
on  scale  A  (cf.  Fig.  15). 

The  instrument  actually  in  use,  the  Mannheim  slide  rule, 
is  a  slight  amplification  of  the  one  just  described  (cf. 
Fig.  16).  It  has  four  scales,  usually  denoted  by  A,  B^  (7,  i>, 
respectively,  the  scales  A  and  D  being  on  the  rule,  and  B 
and  Q  on  the  slide. 


Fia.  16 

The  scale  A  is  composed  of  two  logarithmic  scales  such  as 
that  of  Fig.  14,  so  that  its  right-hand  end  might  be  labeled 
100,  since  log  100  =  2.  On  most  slide  rules,  however,  the 
first  principal  division  on  scale  A  after  9  is  not  labeled  10, 

*  Oughtred's  instruments  were  described  in  publications  of  William  Fos- 
ter, one  of  his  pupils,  in  1632  and  1633. 


64  CALCULATION  BY  LOGARITHMS 

as  in  Fig.  14,  but  1,  the  next  one  is  not  labeled  20,  but  2, 
and  so  on  to  the  last  one,  which  is  again  labeled  1  instead 
of  100  or  10.  Thus,  the  two  halves  of  scale  A  are  exact 
copies  of  one  another.  This  is  done  for  precisely  the  same 
reason  that  the  mantissas  only  are  printed  in  our  tables  of 
logarithms.  The  slide  rule  also  makes  use  of  the  mantissas 
only.  The  characteristics,  or  what  amounts  to  the  same 
thing,  the  position  of  the  decimal  point  in  the  result,  must 
be  obtained  by  inspection  or  by  special  rules. 

Scale  B  is  on  the  upper  edge  of  the  slide,  in  direct  con- 
tact with  scale  A  on  the  rule,  and  is  an  exact  copy  of  scale 
A.  These  two  scales  together  may  be  used  for  multiplica- 
tion and  division  as  explained  above. 

Scale  I)  is  on  the  lower  part  of  the  rule.  It  is  a  single 
logarithmic  scale,  from  1  to  10,  of  the  same  length  as  the 
combined  two  scales  of  A.  The  logarithm  of  any  number 
is  therefore  represented,  on  scale  i>,  by  a  distance  twice  as 
great  as  that  which  represents  the  logarithm  of  the  same 
number  on  scale  A,  It  follows  from  this  that  the  number 
which  is  found  on  scale  A^  vertically  above  any  number  of 
scale  i>,  is  the  square  of  the  latter.  Any  number  on  scale 
i>,  on  the  other  hand,  is  the  square  root  of  the  number  ver- 
tically above  it  on  scale  A. 

Scale  C  is  on  the  lower  edge  of  the  slide,  in  direct  contact 
with  slide  D  on  the  rule.  It  is  an  exact  copy  of  scale  D. 
These  two  scales  together  may  be  used  for  multiplication 
and  division,  according  to  the  same  rules  which  hold  f(5^r 
scales  A  and  B, 

Besides  these  four  scales,  the  slide  rule  is  supplied  with  a 
runner  (cf.  Fig.  16),  which  is  useful  in  performing  com- 
pound operations,  and  also  in  comparing  two  scales  (such  as 
A  and  2)),  which  are  not  in  direct  contact  with  each  other. 
The  runner  was  made  a  permanent  feature  of  the  slide  rule 
by  Mannheim  in  1851.* 

*  Amedee  Mannheim  (1831-1906),  a  distinguished  geometer  of  recent  times. 
The  runner  had  however  been  used  occasionally,  long  before  Mannheim,  by  a 
number  of  English  mathematicians. 


THE  SLIDE  RULE  65 

It  often  happens,  in  manipulating  the  slide  rule,  that  the 
result  is  to  be  sought  opposite  a  number  of  the  slide  which 
falls  outside  of  the  scale  on  the  rule.  In  such  cases,  we  may 
shift  the  slide,  bringing  the  right-hand  index  to  the  place 
which  the  left-hand  index  occupied  previously,  and  read  off 
the  result  as  before.  For,  such  a  shift  has  no  influence  on 
the  mantissa,  since  it  merely  amounts  to  dividing  the  result 
by  10.  On  the  Mannheim  rule,  this  shifting  of  the  slide 
may  be  avoided  by  working  with  scales  A  and  B  rather  than 
with  0  and  D.  Scales  Q  and  i),  however,  have  the  advan- 
tage of  greater  accuracy. 

If  the  slide  be  withdrawn  entirely,  it  will  be  found  to 

have  three  other  scales  on  its  reverse  side,  two  of  which  are 

labeled  S  and  T.      These  are  scales  of  logarithmic  sines  and 

tangents,  respectively,  and  may  be  used  for  calculating  such 

products  as 

c  sin  A^     c  tan  A, 

The  middle  scale  on  the  reverse  side  is  used  for  finding 
the  value  of  the  logarithm  of  a  number,  and  is  important  if 
we  wish  to  compute  a  power  of  a  number  with  a  complicated 
fractional  exponent. 

For  more  complete  information  concerning  the  slide  rule, 
we  must  refer  to  the  manuals  which  are  usually  presented 
to  the  purchaser  of  such  an  instrument.*  Cheap  slide 
rules,  especially  constructed  for  the  beginner,  may  now  be 
obtained  of  all  dealers  under  the  name  Student's  or  Col- 
lege Slide  Rule.  Engineers  and  computers  use  the  slide 
rule  so  extensively  that  the  student  will  find  it  advis- 
able to  make  himself  familiar  with  the  instrument  by  actual 
use. 

The  Mannheim  slide  rule,  which  we  have  described,  admits 
of  three-figure  accuracy.  In  some  (exceptional)  cases, 
results  correct  to  four  decimal  places  may  be  obtained  by  its 
use.  The  Thacher  and  Fuller  slide  rules,  more  compli- 
cated instruments,  but  constructed  on  essentially  the  same 

*  See  also  Raymond's  Plane  Surveying. 


66  CALCULATION   BY  LOGARITHMS 

principles,  admit  of  far  greater  accuracy.  The*  Eichhorn 
Trigonometric  Slide  Rule  was  invented  for  the  purpose  of 
solving  triangles,  and  is  especially  adapted  for  this  work. 
But,  of  course,  it  has  not  the  wide  range  of  usefulness  of  the 
ordinary  slide  rule. 


CHAPTER   VI 

APPLICATION   OF  LOGARITHMS   TO   THE   SOLUTION   OF 
RIGHT   TRIANGLES 

31.  The  general  method.  We  have  shown,  in  Chapter  III, 
how  to  solve  right  triangles  by  means  of  the  natural  functions, 
and  we  have  become  acquainted  with  the  theory  and  use  of 
logarithms  in  Chapter  V.  To  solve  a  right  triangle  by  loga- 
rithms, it  suffices  to  combine  the  results  of  these  two  chapters. 
We  use  the  same  formulae  as  in  Chapter  III,  but  perform 
the  multiplications  and  divisions  by  means  of  logarithms, 
using  the  table  of  logarithmic  sines,  cosines,  etc.,  in  place  of 
the  table  of  natural  functions. 

In  order  to  illustrate  the  various  practical  questions  which 
arise  in  such  a  calculation,  we  shall  give  a  rather  extended 
discussion  of  the  following  example  :  ^ 

Example.  The  legs  of  a  right  triangle  were  found  to  be 
a  =  527.38  feet  and  b  =  621.24  feet.  Calculate  the  hypote- 
nuse and  the  acute  angles  A  and  B, 


32.    The 

'rolution. 

i     preliminary     graphic 

We  first  make  a   drawing,   approxi- 
mately to  scale,  making 

a  =  5.3  centimeters, 
b  =  6.2  centimeters. 

We  find  by 

neasurement 

c 

A 

=  8,1  centimeters, 
=  40°.  5,   5  =  49°.5. 

This  figure  and  these  measurenaents  serve  two  purposes. 
In  the  first  place,  the  figure  helps  us  pick  out  the  formulae 
which  we  shall  need  for  the  trigonometric  solution  of  our 
triangle.  In  the  second  place,  comparison  of  the  approxi- 
mate values  of  the  unknown  quantities  obtained  graphically, 

67 


68     SOLUTION  OF  RIGHT   TRIANGLES   BY  LOGARITHMS 

with  the  final  results  as  obtained  by  calculation,  constitutes 
a  valuable  check.  If  the  results  obtained  by  the  two  methods 
should  differ  by  more  than  can  be  accounted  for  by  the  in- 
accuracies of  the  graphic  solution,  we  must  look  for  a  mis- 
take in  our  calculation. 

33.  The  gross  errors.  Mistakes,  which  are  large  enough 
to  be  detected  by  means  of  a  graphic  check,  are  known  as 
gross  errors  and  are  usually  due  to  one  of  the  following 
causes : 

1.  The  lise  of  a  wrong  formula. 

2.  A  misplaced  decimal  point  or,  what  amounts  to  the 
same  thing,  an  erroneous  characteristic. 

3.  The  use  of  a  number  taken  from  a  wrong  column  in 
the  tables,  resulting,  for  instance,  in  erroneously  using  the 
log  cos  of  an  angle  in  place  of  the  log  sin. 

4.  Addition  of  two  logarithms  when  subtraction  is  re- 
quired or  vice  versa. 

5.  Purely  arithmetical  errors  of  addition  and  subtraction. 
The  errors  of  the  first  four  classes  can  be  avoided  by  the 

exercise  of  a  sufficient  amount  of  care.  The  student  should 
not  attempt  to  gain  speed  in  calculation  until  he  has  first 
learned  to  be  accurate.  The  errors  of  the  last  class  are 
quite  unavoidable,  but  they  will  usually  be  detected  almost 
as  soon  as  made  if  the  ordinary  arithmetical  checks  for 
addition  and  subtraction  be  applied  every  time  that  one  of 
these  operations  is  used. 

34.  Selection  of  formulae  and  checks.  After  completing 
the  approximate  graphical  solution  of  a  triangle,  we  pick  out 
the  formulae  which  we  wish  to  use  in  the  computation. 

In  our  example,  these  are  the  following  : 

(1)  tan^  =^,  c  =  -^— =  — ^,  5  =  90°-^, 

6  sin  ^      cos  A  d 

or,  in  logarithmic  form, 

(2)  log  tan  ^  =  log  a  -  log  J,  5  =  90°  -  J, 
log  c  =  log  a  —  log  sin  A  =  log  h  —  log  cos  ^ . 

We  have  two  formulas  for  c,  and  in  most  cases  it  makes  little  differ- 
ence which  one  we  decide  to  use.     If  we  were  to  use  both,  the  agi'eement 


¥ 


SKELETON  FORM   OF   SOLUTION  69 


of  the  two  results  would  constitute  a  partial  check  on  the  accuracy  of  our 
work.  It  would  not  be  a  total  check  however ;  a  mistake  in  the  loga- 
rithm of  a  or  6  could  not  be  detected  by  means  of  it.*  It  is  hardly 
worth  while  therefore  to  use  both  formulae  for  c.  That  one  is  to  be  pre- 
ferred which  has  the  greater  denominator,  as  the  result  obtained  from  it 
is  likely  to  be  the  more  accurate. 

For  a  complete  check,  we  may  make  use  of  the  equation 
a^+b^  =  c2, 
which,  however,  we  prefer  to  write  in  the  form 
(3)  a  =  V^^-^Tp  =  V(6--|-&)  (c-b), 

which  is  more  convenient  for  logarithmic  computation. 

It  should  not  be  necessary  to  write  the  formulae  in  the  logarithmic 
form  (2).  Form  (1)  is  shorter  and  more  directly  connected  with  the 
geometry  of  the  problem.  Moreover,  it  contains  all  of  the  information 
that  is  necessary  for  the  solution  of  the  problem,  for  anybody  who  has 
studied  logarithms. 

||  35.  The  framework  or  skeleton  form.  Having  selected 
the  formulae,  we  proceed  to  plan  the  details  of  the  computa- 
tion by  providing  a  definite,  properly  marked  place  for  every 
number  which  will  be  needed  in  the  course  of  the  work. 
Moreover,  we  shall  plan  these  details  in  such  a  way  that  those 
numbers  which  are  to  be  combined  by  addition  or  subtraction 
will  have  their  places  in  the  same  vertical  column  next  to 
each  other. 

In  our  particular  example  we  may  adopt  the  following  framework : 


k 


Given  {    ;: 

(1)                               log&  = 

(4) 

(2)                       log  cos  A  = 

(7) 

loga  = 

(3)                               logc  = 

(4) -(7)  =  (8) 

log&  = 

(4)                                     c  = 

(9) 

log  tan  A  = 

(3) -(4)  =  (5)                  b= 

(2) 

(A  = 

(6)                                c-b  = 

(9) -(2)  =  (10) 

Results  i  B  = 

(16)                              c  +  b  = 

(9) +  (2)  =  (11) 

.  c  = 

(9)                      log(c-b)  = 

(12) 

log  (c  +  b)  = 

(13) 

l0g(c2-&2)=: 

(12) +  (13)  =  (14) 

Check  ]l«g^(^^-^^)  = 
I'                  log  a  = 

1(14)  =  (15) 

(3) 

*  Since  such  a  mistake  could  be  interpreted  as  leading  to  the  correct  solution 
of  a  triangle  different  from  the  given  one,  namely,  that  one  whose  sides  a'  and  b' 
have  as  logarithms  the  values,  which  were,  by  mistake,  assigned  to  a  and  b. 


70     SOLUXrON  OF   RIGHT   TRIANGLES   BY   LOGARITHMS 


The  numbers  in  parenthesis  merely  indicate  the  order  in  which  this 
skeleton  form  may  be  filled  in,  and  how  some  of  the  results  are  obtained. 
The  student  should  use  these  numbers  only  to  aid  him  in  understanding 
the  construction  of  the  framework  and  the  plan  of  the  computation. 
They  should  not  be  used  in  writing  out  the  actual  calculation. 

36.  The  computation.  We  are  now  prepared  to  carry  out 
the  computation.  This  should  be  done  on  paper  ruled  into 
squares  of  such  a  size  that  each  figure  may  conveniently 
occupy  one  square.     We  obtain  the  following  results  : 


Given 


a  =  527.38 
h  =  621.24 
log  a  =  2.72212 
log  6  1:^2.79326 
log  tan  ^  =  9.92886-10 
[  ^  =  40°  19'.69 
Results  \  B=^r  40'.31 
c  =  814.92 


logZ>  =  2.79326 

log  cos  ^  =  9.88215  -  10 

logc  =  2.91111 

c=    814.92 

b=    621.24 

c-b=    193.68 

c  +  b  =  1436.16 

log  (c- 6)  =2.28709 

log  (c  +  &)  =  3.15720 

log  (c2  -  62)  =  5.44429 


logVc2-  &2  =  2.722151 


Check. 


loga  =  2.72212] 

Remark.  We  observe  that  the  check  is  not  absolute.  The  agreement 
is  as  close,  however,  as  we  should  expect.  The  inevitable  inaccuracies, 
arising  from  the  neglected  higher  decimal  places,  often  manifest  them- 
selves by  discrepancies  amounting  to  several  units  of  the  fifth  decimal 
place.     Consequently,  we  may  declare  the  check  to  be  satisfactory.* 

37.  Revision  of  the  computation  when  the  check  is  un- 
satisfactory. If,  in  the  solution  of  such  an  example,  the 
results  fail  to  check  satisfactorily,  the  magnitude  of  the  dis- 
crepancy will  help  us  to  locate  the  error.  If  the  discrepancy 
is  very  great,  the  error  must  be  one  of  the  gross  kind  which 
we  have  discussed  in  Art.  33.  In  case  of  a  comparatively 
small  discrepancy,  our  error  is  probably  due  to  one  of  the 
following  causes : 

1.  Purely  arithmetical  errors  of  addition  and  subtraction 
in  the  last  few  decimal  places. 


*  If   a   and    6   differ   considerably,   use  as  check   a  =  v(c  —  6)  (c  +  b)    or 

6  =  Vic  —  a){c-{-a),  according  as  6  < a  or  &  >  a. 


EXERCISES  ON   RIGHT   TRIANGLES  71 

2.  Inexact  interpolation,  which  would  ordinarily  affect 
only  the  last  decimal  place. 

3.  Addition  of  the  correction  obtained  by  interpolation 
when  it  should  be  subtracted,  or  vice  versa. 

This  last  mistake  may  be  avoided  by  carefully  inspecting 
the  table  after  the  interpolation  has  been  completed,  so  as  to 
make  sure  that  the  quantity  calculated  actually  lies  between 
the  two  numbers  of  the  table  between  which  it  should  fall. 

EXERCISE  XV 

In  each  of  the  following  examples  (1-10),  two  parts  of  a  right 
triangle  are  given  in  the  usual  notation.     Find  the  other  parts : 

1.  c  =  627,        A  =  23°  30'.  6.   a  =  13.690,     b  =  16.926. 

2.  c  =  934,        B  =  76°  25'.  7.   a  =  67.291,  c  =  110.970. 

3.  a  =  637,        A  =  4:°  35'.  8.    6  =  618.42,  c  =  1843.70. 

4.  &  =  48.532,  B  =  36°  44'.00.  9.   a  =  965.24,  A  =  75°  15'.2. 

5.  a  =  38.313,    h  =  19.522.  10.   a  =  7.3298,  b  =  6.1032. 

An  isosceles  triangle  may  be  divided  into  two  equal  right  triangles 
by  dropping  a  perpendicular  from  the  vertex  to  the  base.     Using  the 
notations  of  Fig.  18,  find  the  missing  sides  and  angles 
of  the  following  isosceles  triangles.     (Exs.  11-13.) 

11.  b  =  2.1452,       B  =  121°  14'.60. 

12.  A  =  52°  10'.2,    a  =  600.20. 

13.  h  =  7.447,        A  =  76°  14'.00. 

14.  Prove  that  the  area  S  oi  a  right  triangle  is 

S=  ^bc  sin  A  =  ^  ac  cos  A . 

15.  Prove  that  the  area  S  oi  a  right  triangle  is 

S  =  I  c^  sin  A  cos  A . 

16-25.   Find  the  area  of  each  of  the  right  triangles  in  Exs.  1-10. 

26-30.  Find  formulae  for  the  area  of  the  isosceles  triangle  of  Fig.  18, 
in  terms  of  b  and  h ;  a  and  b ;  a  and  Ji ;  a  and  B ;  a  and  A. 

31-33.  Apply  the  results  of  Exs.  26-30  to  find  the  areas  of  the  isosceles 
triangles  of  Exs.  11-13. 

34.  Show  that  the  perimeter  jt?  of  a  regular  polygon  of  n  sides  in- 
scribed in  a  circle  of  radius  R  is 

jo  =  2ni2sini^, 
n 


n 


72     SOLUTION   OF   RIGHT   TRIANGLES  BY  LOGARITHMS 

that  the  radius  r  of  the  inscribed  circle  is 

and  that  the  area  S  of  the  polygon  is 

5=wi22sin— cos 

n  n 

35.  Find  the  radius  of  the  inscribed  circle,  the  perimeter  and  the  area 
of  a  regular  pentagon,  if  the  radius  of  the  circumscribed  circle  is  12  feet. 

36.  Find  the  perimeter,  the  length  of  one  side,  the  radii  of  the  in- 
scribed and  circumscribed  circles  of  a  regular  octagon  whose  area  is  24 
square  feet. 

37.  Since  the  polygon  of  Ex.  33  approaches  the  circle  of  radius  R  as 

1 80*^         1  SO'' 

limit  when  n  grows  beyond  all  bound,  what  limits  do  n  sin cos 

n  n 

and  2  n  sin ,  respectively,  approach  ? 

n 

38.  Applications  to  simple  problems  of  surveying,  navigation, 
and  geography.  The  connection  between  surveying  and 
trigonometry  is  so  obvious  as  to  require  no  further  explana- 
tion. Moreover,  we  have  already  discussed  this  relation  in 
Chapter  1. 

Many  of  the  following  examples  are  concerned  with  simple 
problems  of  surveying,  and  most  of  the  technical  terms  which 
occur  in  them  are  self-explanatory.  Nevertheless,  we  shall 
give  a  brief  discussion  of  these  terms,  so  as  to  make  the  appli- 
cations seem  more  concrete  and  vivid.  The  student  who 
wishes  to  know  more  about  the  subject  should  consult  a 
treatise  on  surveying.* 

A  pltmib  line  is  a  cord  to  one  end  of  which  is  attached  a 
weight.  If  such  a  plumb  line  be  suspended,  by  fastening  the 
other  end  of  the  cord  to  a  fixed  support,  it  will  oscillate  to 
and  fro,  and  finally  come  to  rest  in  its  position  of  equilibrium, 
which  is  called  the  vertical  line  of  the  place  of  observation. 

Since  the  earth  is  approximately  spherical  in  shape  and  the 
plumb  line  points  toward  the  earth's  center,  the  vertical  lines 
of  different  places  are  not  parallel.     But  the  angle  between 


♦For  instance,  Raymond's  Plane  Surveying. 


PROBLEMS   OF   SURVEYING  AND  NAVIGATION 


73 


the  vertical  lines  of  two  stations  which  are  not  very  far 
apart  (say  ten  miles),  is  so  small  that  for  most  purposes  these 
lines  may  be  regarded  as  parallel.  When  a  tract  of  land  (to 
be  surveyed)  is  comparatively  small,  it  is  therefore  legitimate 
to  neglect  the  effect  of  the  earth's  curvature,  and  the  problem 
becomes  one  of  plane  survejring.  The  more  difficult  problems 
connected  with  a  geodetic  survey^  in  which  the  earth's 
spherical  form  is  taken  into  account,  require  knowledge  of 
the  methods  of  spherical  trigonometry. 

We  are  concerned  with  plane  surveying  only,  so  that  we 
shall  regard  the  vertical  lines  of  all  places  which  occur  in 
such  a  survey  as  parallel. 

A  vertical  plane  is  one  which  contains  a  vertical  line. 

A  horizontal  line  or  plane  is  one  which  is  perpendicular  to 

ta  vertical  line. 
I  An  inclined  line  or  plane  is  one  which  is  neither  vertical 
nor  horizontal. 

An  angle  is  said  to  be  a  horizontal  or  vertical  angle,  accord- 
ing as  the  plane  of  its  sides  is  a  horizontal  or  vertical  plane. 
Both  horizontal  and  vertical  angles  may  be  measured  by 
means  of  the  transit  (Art.  2).  Inclined  angles  may  be 
measured  by  means  of  an  instrument  known  as  a  sextant. 


Horizontal  line 


D=rAngle  of  depression 
Fig.  19 


Horizontal  line 
JS7=AngIe  of  elevation 
Fig.  20 


■  The  angle  which  the  line  of  sight  from  the  observer  to  an 
object  makes  with  a  horizontal  line,  in  the  same  vertical 
plane,  is  called  the  angle  of  elevation  or  the  angle  of  depres- 


74      SOLUTION   OF   RIGHT   TRIANGLES  BY  LOGARITHMS 


Fig.  21 


sion,  according  as  the  object  is  above  or  below  the  horizontal 
plane  of  the  observer  (cf.  Figs.  19  and  20). 

The  angle  subtended  by  a  line  is  that  which  is  obtained  by 

joining  the  extremities  of  the  line  to  the  eye  of  the  observer.. 

The  direction  or  bearing  of  any  horizontal  line  is  usually 

described  by  means  of  the  angle  which  it  makes  with  the 

north-south  line  or  meridian,  the 
latter  being  located  approximately 
with  the  help  of  a  surveyor's 
compass.  Surveyors  always  meas- 
ure the  bearing  of  a  line  as  an 
acute  angle  from  the  north  or  south 
end  of  the  meridian  toward  the 
east  or  west  point,  as  the  case  may 
be.  Thus,  in  Fig.  21,  the  bearing 
of  OA  is  N  45°  E,  that  of  OB  is  S 
10°  W,  that  of  0(7  is  N  30°  W. 
Surveyors  usually  measure  dis- 
tances by  means  of  a  Gunter's  chain,  which  is  4  rods  or  6Q 
feet  long,  and  is  divided  into  100  links.  For  this  reason,  the 
operation  of  measuring  the  length  of  a  line  in  the  field  is 
frequently  called  chaining. 

In  order  to  measure  the  difference  of  level  between  two 
places,  A  and  B  (cf.  Fig.  22),  the  observer  at  0  first  makes 
his  telescope  point  in  a  horizontal  direction  by  means  of  a 
spirit  level  attached  to  the  telescope.  An  assistant  holds  a 
graduated  rod,  i2,  in  a  vertical  position  at  A,  and  the  ob- 
server at  0  reads,  by 
means  of  the  telescope,  the 
division  on  the  graduated 
rod  where  it  is  struck  by 
the  horizontal  line  of  sight. 
He  repeats  this  operation  with  the  rod  at  B.  The  difference 
between  the  two  readings  gives  the  difference  of  level  between 
A  and  B.  Of  course,  if  the  difference  of  level  between  A  and  B 
exceeds  the  length  of  the  rod,  intermediate  stations  must  be 
introduced.     This  operation  is  known  technically  as  leveling. 


Fig.  22 


PROBLEMS   OF   SURVEYING  AND   NAVIGATION        75 

The  navigator   does  not  always  express  bearings  in  the 
same  language  as  the  surveyor.       He   divides  the  circum- 
^    =    ^.  ference  into  32  equal  parts, 

called  points  of  the  com- 
pass. Thus  one  point  of 
the  compass  is  an  angle  of 
llj  degrees.  The  division 
points   are  named,  as  indi- 


FiG.  23.  —  The  points  of    the    Mariner's 
Compass 

cated  in  Fig.  23,  with  obvious  reference  to  the  four  cardinal 
points  of  the  compass,  —  north,  south,  east,  and  west. 

The  navigator  also  makes  use  of  the  terms  departure  (to 
denote  the  east  and  west  component  of  ,a  course),  and  differ- 
ence in  latitude  (to  denote  the  north  and  south  component). 
These  terms  are  illustrated  in  Fig.  24. 


EXERCISE    XVI 

In  solving  the  following  problems,  the  student  should  exercise  his 
judgment  in  regard  to  the  number  of  decimal  places  to  be  used  in  the 
calculation.  (See  Chap.  I,  Arts.  1,  2;  Chap.  IV,  Arts.  14,  15.)  Many 
of  these  problems  may  be  solved  by  means  of  three  place  tables  (Tables 
IX,  X,  and  XI  of  our  collection),  or  by  means  of  the  slide  rule.  (See 
Chap.  V,  Arts.  29,  30.)  In  all  of  the  examples  the  slide  rule  may  be 
used  as  a  check. 

1.  At  a  point  180.00  feet  away  from  the  base  of  a  tower  and  in  the 
same  horizontal  plane  with  it,  the  angle  of  elevation  of  the  top  was  found 
to  be  65°  40'.5.     Find  the  height  of  the  tower. 

2.  From  the  top  of  a  cliff  120  feet  above  the  level  of  a  lake,  the  angle 
of  depression  of  a  boat  was  found  to  be  27°  40'.  What  is  the  air  line 
distance  from  the  top  of  the  cliffc'  to  the  boat? 


76      SOLUTION   OF   RIGHT   TRIANGLES   BY  LOGARITHMS 

3.  In  order  to  measure  the  width  of  a  river,  a  base  line  AC  is  meas- 
ured along  one  bank  215.6  feet  long.  By  means  of  a  transit,  a  point  B 
is  located  on  the  opposite  bank  such  that  ACB  is  a  right  angle.  The 
angle  BAC  is  found  to  be  55°  16'. 2.  What  is  the  width  BC  of  the 
river  ? 

4.  From  the  top  of  a  mountain  2653  feet  above  the  floor  of  the 
valley,  the  angles  of  depression  of  two  farmhouses  in  the  level  valley 
beneath,  both  of  which  were  due  east  of  the  observer,  were  found  to  be 
25°  and  56°.     What  is  the  horizontal  distance  between  the  two  houses  ? 

5.  From  the  top  of  a  hill,  the  angles  of  depression  of  two  consecu- 
tive milestones  on  a  straight  level  road,  running  due  south  from  the  ob- 
server, were  found  to  be  22°  31'  and  48°  15'.     How  high  is  the  hill? 

Hint.     Treat  this  problem   as   one   involving  two   unknowns:  1st, 
the  height  of   the  hill;   2d,   the   horizontal  distance  from   one   of  the 
milestones  to  the  foot  of  the  perpendicular  dropped  from  the  top  of  the 
hill  to  the  horizontal  plane  of  the  road. 

6.  Three  lighthouses  A,  B,  C,  are  situated  as  in 
Fig.  25,  the  triangle  ABC  being  right-angled  at  A.  At 
the  moment  when  a  ship  S  is  crossing  the  line  AC,  the 
angle  A  SB  is  found  to  be  75°.  If  the  distance  between 
the  lighthouses  A  and  B  is  12  miles,  what  are  the  dis- 
tances from  S  to  A  stwd.  B'i 

7.  A  light  on  a  certain  steamer  is  known  to  be  35 
Fig.  25          feet  above  the  water.      An  observer  on  the  shore,  whose 

instrument  is  5  feet  above  the  water,  finds  the  angle  of 
elevation  of  this  light  to  be  5°.  What  is  the  distance  from  the  observer 
to  the  steamer? 

8.  What  angle  does  a  mountain  slope  make  with  a  horizontal  plane, 
if  it  rises  200  feet  in  a  horizontal  distance  of  one  tenth  of  a  mile  ? 

9.  The  cable  of  a  captive  balloon  is  835  feet  long.  Assuming  the 
cable  to  be  straight,  how  high  is  the  balloon  when  all  of  the  cable  is  out 
if,  owing  to  the  wind,  the  cable  makes  an  angle  of  25°  with  a  vertical 
line  ?  s^ g 

10.  A  ship  is  sailing  due  west  at  the  rate  of  8.9  miles 
per  hour.  A  lighthouse  is  observed  due  south  at  10  p.m. 
The  bearing  of  the  same  lighthouse  at  11 :  55  p.m.  was 
S.  34°  E.  Find  the  distance  from  the  lighthouse  to  the 
ship  at  the  time  of  the  second  observation. 

Hint.  In  Fig.  26,  <S  and  S'  represent  the  two  positions 
of  the  ship  and  L  represents  the  lighthouse.  Angle 
SS'L  =  90°  -  34°.  Fia.  26 


i 


APPLICATIONS  INVOLVING  RIGHT   TRIANGLES       77 

11.  Find  the  area  of  the  tract  of  land  corresponding  to  the  following 
description.  From  A  the  boundary  line  runs  N.  24°  E.  20  chains  to  By 
thence  N.  85°  W.  35.67  chains  to  C,  and  thence  back  to  A. 

12.  The  shadow  of  a  chimney,  50  feet  high,  is  60  feet  long.  What  is 
the  altitude  (or  angle  of  elevation)  of  the  sun  at  that  instant  ? 

13.  The  last  row  of  seats  in  a  circular  tent  is  20  feet  away  from  the 
central  pole,  which  is  18  feet  high,  and  which  is  to  be  fastened  by  ropes 
from  its  top  to  stakes  driven  in  the  ground.  How  long  must  these  ropes 
be  in  order  that  they  may  be  6  feet  above  the  ground  over  the  last  row 
of  seats,  and  at  what  distance  from  the  center  must  the  stakes  be 
driven  ? 

14.  How  long  must  a  ladder  be  to  reach  a  window  45  feet  high,  di- 
rectly above  a  porch  15  feet  high,  if  the  porch  projects  10  feet  from  the 
building? 

15.  A  building  125  feet  high,  with  a  flat  roof,  faces  north  on  a  boule- 
vard. The  distance  from  this  building  to  the  one  directly  opposite  is 
180  feet.  How  far  back  from  the  edge  of  the  roof  should  a  chimney  6 
feet  high  be  placed,  so  as  to  be  invisible  from  any  point  on  the  boule- 
vard due  north  of  the  chimney  ? 

16.  A  cylindrical  pipe  36  inches  in  diameter  is  to  be  joined  to  a 
second  cylindrical  pipe  18  inches  in  diameter.  The  axes  of  the  two 
cylinders  are  pieces  of  the  same  horizontal  line  and  their  ends  are  6  feet 
apart.  The  joining  piece  is  to  be  in  the  form  of  a  frustum  of  a  cone. 
Draw  a  sectional  view  of  the  joining  piece  and  compute  the  length  of 
the  slanting  side  and  the  angles. 

17.  We  wish  to  construct  a  house  with  a  gable  roof.  If  the  house  is 
25  feet  wide,  if  the  height  under  the  eaves  is  27  feet  and  the  height  to 
the  ridge  pole  35  feet,  how  long  must  the  rafters  be  so  that  their  ends 
may  be  at  a  horizontal  distance  of  2  feet  from  the  side  of  the  house  ? 

18.  The  angle  of  elevation  of  the  center  of  a  spherical  balloon  20  feet 
in  diameter  was  found  to  be  65°.  The  angle  which  it  subtended  at  the 
same  time  was  2°  30'.  What  is  the  height  of  the  balloon  above  the 
horizontal  plane  of  the  observer? 

19.  Two  stations,  A  and  B,  are  to  be  connected 
by  a  railroad.  Both  stations  are  in  the  same  hori- 
zontal plane,  and  5  is  35  miles  northeast  of  A.  The 
two  stations  are  separated  by  a  lake,  which  termi- 
nates at  a  point  C,  12  miles  north  and  29  miles  east 
of  A.  Find  the  lengths  and  bearings  of  the  two 
portions  of  the  road  from  ^  to  C  and  from  C  to  B. 

20.  A  lecture  room,  50  feet  long  and  18  feet  high,  is  to  be  supplied 
with  a  sloping  floor.     The  front  part  of  the  floor,  for  the  first  ten  feet,  is 


78     SOLUTION   OF   RIGHT   TRIANGLES   BY   LOGARITHMS 

to  be  horizontal  so  as  to  admit  of  the  placing  of  lecture  tables  and  ap- 
paratus. The  highest  part  of  the  sloping  floor,  at  the  back  of  the  room, 
is  to  be  8  feet  from  the  ceiling.  What  length  of  sloping  timbers  is  re- 
quired for  this  construction?  Each  of  these  timbers  is  to  be  supported  at 
both  ends  and  by  six  intermediate  uprights  placed  at  equal  horizontal 
distances  from  each  other  and  from  the  end  supports.  How  far  should 
each  of  these  eight  supports  project  above  the  horizontal  part  of  the 
floor? 

21.  In  order  to  determine  the  height  of  a  mountain  above  a  level 
plane,  we  may  measure  a  horizontal  base  line  of  length  h  in  the  same 
vertical  plane  with  the  summit  of  the  mountain  and  observe  the  angles  of 
elevation,  A  and  B,  of  the  summit  from  the  two  ends  of  the  base  line. 
Find  a  formula  for  the  vertical  height  h  of  the  mountain  above  the  level 
of  the  plane. 

22.  Apply  the  formula  of  Ex.  21  to  the  case  h  =  100  feet,  A  =  30°, 
B  =  35°. 

23.  A  flagstaff,  known  to  be  h  feet  in  length,  stands  on  top  of  a  cliff. 
An  observer,  in  the  same  horizontal  plane  with  the  base  of  the  cliff,  finds 
the  angles  of  elevation  of  the  top  and  bottom  of  the  flagstaff  to  be  A  and 
B  respectively.     Find  a  formula  for  the  height  of  the  cliff. 

24.  Apply  the  formula  of  Ex.  23  to  the  case  A  =25,  ^  =  40=  25',  B  = 
37°  10'. 

25.  The  angle  of  elevation  of  the  top  of  a  spire  from  the  third  floor  of 
a  building  was  35°  10'.  The  angle  of  elevation  from  a  point  directly 
above,  on  the  fifth  floor  of  the  same  building,  was  25°  33'.  What  is  the 
height  of  the  spire  and  its  horizontal  distance  from  the  place  of  observa- 
tion, if  the  distance  between  consecutive  floors  is  12  feet  and  the  first  floor 
rests  on  a  basement  5  feet  above  the  level  of  the  street  ? 

26.  Prove  the  following  statement.     If  R  is  the  radius  of  the  earth 
regarded  as  a  sphere,  the  radius  of  the  parallel  of  lati- 
tude which  passes  through  a  place  P  of  latitude  L,  is 

r  =  R  cos  L. 

Hint.  Use  Fig.  28,  where  0  denotes  the  earth's 
center,  N  and  S  the  north  and  south  poles,  OE  =  R 
the  radius  of  the  equator,  Z  EOP  =  L  the  latitude  of 
the  place  P,  and  r  =  MP  the  radius  of  the  parallel 
of  latitude  which  passes  through  P. 

27.  Let  d  be  the  length,  in  miles,  of  a  degree  of 

longitude  at  the  equator.  Show  that  the  length  of  a  degree  of  longi- 
tude, at  latitude  L,  will  be  d  cos  L. 


APPLICATIONS  INVOLVING   RIGHT   TRIANGLES 


79 


28.  Show  that  the  radius  r  (in  feet)  of  the  horizon  of  an  observer,  h 
feet  above  the  earth's  surface,  is  given  by  the  formula 

if  R  denotes  the  earth's  radius  expressed  in  feet. 

Hint.  Use  Fig.  29,  where  OQ  =  OM  =  R, 
MP  =  h,  QN  =  r,  using  the  angle  NOQ  =  ^  as 
auxiliary. 

29.  A  micrometer  screw  is  to  be  cut   from    a 

cylindrical  steel  rod,  5  millimeters  in  diameter,  in  such  a  way  that  one 
complete  revolution  of  the  screw  will  move  the  wire 
W,  attached  to  the  movable  frame  F  (Fig.  30), 
through  a  distance  of  one  millimeter.  What  angle 
will  the  thread  of  the  screw  make  with  a  plane  per- 
pendicular to  the  axis  of  the  screw  ? 

30.  A  street  railway  track  is  d  feet  from  the  curbstone.  In  passing  a 
corner  (Fig.  31),  where  the  street  is  deflected  through  an  angle  of  K"^,  it 
is  desired  to  have  the  rail  pass  at  a  distance 
of  d'  feet  from  the  corner.  Show  that  the 
radius  of  the  circular  curve  A  CB  must  be 

d-d'  cos  — 
2 


Fig.  30 


1  -  cos  - 
2 

31.  Solve  the  problem  of  Ex.  30  numeri- 
cally for  the  cases  d  =  10  feet,  rf'  =  4  feet, 
K=90°:  and  rf  ==  9  feet,  d'  =  3.5  feet,  K  =  60°. 


Fig.  31 


32.   In  order  to  find  the  horizontal  distance  between  the  points  A  and 
F  which  are  at  diiferent  levels  and  situated  on  opposite  sides  of  a  rolling 

valley  (see  Fig.  32),  the  dis- 
tances AB  —  c?i,  BC  =  d^ 
...  EF  =  d^ are  measured  along 
the  ground  by  chaining.  A 
transit  is  placed  at  0,  and 
A'B'C  •••  F'  represents  the 
line  of  sight  of  the  instrument.  This  line  of  sight  is  made  horizontal 
by  means  of  a  spirit-level  attached  to  the  telescope.  The  vertical 
distances 

AA'  =h.,BB'  =Ao,  CC 


i„  EE'  =  h„ 


FF'  =  7i, 


are  measured  by  a  rod.     (Cf.  Fig.  22  for  method  of  using  rod.) 
Show  that  the  distance 

A  'F'  =  c?j  cos  i\  +  ^2  cos  ic^  +  c?3  cos  Zg  +  d^  cos  i^  +  d^  cos  i^, 


80     SOLUTION  OF  RIGHT   TRIANGLES  BY   LOGARITHMS 


where  i\,  i^  ig,  i^,  i^  are  the  angles  of  inclination  of  AB,  BC,  CD,  DE,  EF, 


respectiv^ely. 


The  angles 


The  angle  ij  is  determined  by  the  equation 
sin  ,•  =  ^2-^. 


{5  are  determined  in  similar  fashion. 


33.   In  constructing  a  telegraph  line  across  a  hill  ABC,  •••  /  (Fig.  33), 
posts  were  set  at  ^,  jB,  C,  •••  /,  these  points  being  determined  by  level 

chaining  *  in  such  a  way  that 
the  horizontal  distance  be- 
tween any  two  of  them  A'B'  = 
B'C  =  CD'  =  ...  =  HT  =  d 
feet.  By  leveling,  the  eleva- 
tions of  the  points  A,B,  C,  etc., 
were  foUnd  to  be 

A  A'  =\,  BB'  =  \, 
CC<  =  ho.  •••  //' 


B'      C 


G' 


I' 


D'     E'      F' 

Fig.  33  _ 

Find  a  method  for  computing  the  amount  of  wire  required  between  A 
and  I,  assuming  that  the  telegraph  poles  are  vertical  and  of  the  same 
height,  and  making  no  allowance  for  sag. 

39.  Right    triangles  of    unfavorable   dimensions.      If    the 

hypotenuse  and  one  side  (say  h  and  c)  are  given,  we  have 
the  equation 

(1)  cos  A=~ 


to  determine  the  angle  A.  But,  if  h  differs  very  little  from 
<?,  the  value  of  cos  A  will  be  very  close  to  unity  and,  as  we 
observed  in  Art.  28,  it  will  be  impossible  to  determine  A  with 
any  degree  of  accuracy  from  this  equation. 

A  surveyor  will  usually  (not  always)  be  in  a  position  to 
avoid  this  difficulty.  For  he  has  a  certain  amount  of  liberty 
in  the  choice  of  his  triangles.  But,  in  many  problems  of 
astronomy  and  mathematical  geography,  no  such  choice  is 
possible,  so  that  it  becomes  a  matter  of  practical  importance 
to  find  a  formula  for  determining  the  angle  J.,  which  shall 
not  be  liable  to  the  same  objection  as  (1). 


*In  level  chaining,  the  surveyor's  chain  or  tape  is  held  in  a  horizontal  position, 
so  as  to  measure  the  horizontal  distance  between  two  points  and  not  the  distance 
along  the  slope. 


APPLICATIONS   INVOLVING  RIGHT   TRIANGLES        81 

Such  a  formula  may  be  obtained  as  follows.  In  Fig.  34, 
draw  AD^  the  bisector  of  the  angle  A^  and  also  BE  perpen- 
dicular to  AD.     Then 

1  J.  =  Z  QAB  =  Z  CBE 

(both     angles    being    complementary    to 
Z  AEB),  and  1 

AE  =  AB  =  c,   CE  =  c  -  h. 

Consequently  we  find,  from  the  right  triangle  BCE, 

But 


a  =  Vc2  -  52  =  V(c  -  ^)(^  +  ^), 
so  that  we  may  write,  in  place  of  (2), 


(3)  tan^^  =  ^ 


c  +  b 


This  is  the  desired  formula,  which  should  be  applied  in- 
stead of  (1),  whenever  the  value  of  b  is  very  close  to  that  of 
c,  i.e.  whenever  the  angle  A  is  very  small. 

Some  of  the  following  examples  will  illustrate  the  useful- 
ness of  this  formula  as  well  as  the  application  of  Table  III 
for  the  functions  of  small  angles. 

EXERCISE  XVII 

1.  At  what  distance  may  a  mountain  14,000  feet  high  be  seen  at  sea, 
if  the  earth's  radius  is  3963  miles  ? 

2.  How  high  above  the  earth's  surface  must  a  balloon  rise,  in  order  to 
enable  an  observer  to  see  a  point  50  miles  away? 

3.  If  the  moon's  parallax  (the  angle  which  the  earth's  radius  sub- 
tends as  seen  from  the  moon)  is  57 ',  and  if  the  earth's  radius  is  3963 
miles,  what  is  the  moon's  distance  from  the  earth  ? 

4.  If  the  angular  diameter  of  the  moon,  as  seen  from  the  earth,  is 
31'  20"  and  the  distance  from  the  earth  to  the  moon  is  239,100  miles, 
what  is  the  moon's  diameter  in  miles? 

5.  If  the  distance  from  the  earth  to  the  sun  is  92,000,000  miles,  and 
the  angular  diameter  of  the  sun  as  seen  from  the  earth  is  32',  what  is 
the  diameter  of  the  sun  in  miles  ? 


CHAPTER   VII 

THEORY   OF   OBLIQUE  TRIANGLES 

40.  The  area  of  an  oblique  triangle  in  terms  of  two  of  its 
sides  and  the  included  angle. 

Let  the  sides  ^,  c  of  a  triangle  and  the  angle  A  be  given. 

If  we  consider  the  side  AB  =  <?  as  the  base,  then  the  alti- 
tude CD=h  is  the  length  of  the  perpendicular  dropped  from 
C  to  AB.  The  foot  I>  of  this  perpendicular  may  fall  on  the 
c 


A  c  D         B         .  A       c     B        D 

Fig.  35  Fig.  36  Fig.  37 

line  segment  AB  (Fig.  35),  to  the  right  of  B  (Fig.  36),  or 
to  the  left  of  A  (Fig.  37).     In  all  of  these  cases  we  have, 

(1)  8=lch, 
if  S  denotes  the  area  of  the  triangle. 

Now  h  can  be  expressed  in  terms  of  the  given  quantities, 
5,  c,  and  A^  as  follows. 

In  Figs.  35  and  36,  in  which  A  is  an  acute  angle,  we  have 

-■=  ^inA  ov  h  =  h  sin  A^ 

0 

and  therefore,  by  substituting  this  value  of  h  in  (1), 

(2)  aS'  =  ^  he  sin  A. 

If  A  is  an  obtuse  angle  (Fig.  37),  we  have 

^  =  sin  Z  7)^(7  =  sin  (180°  -  ^),  or  ^  =  5  sin  (180°  -  A\ 
h 

and  consequently,  by  substituting  this  value  of  ^  in  (1),- 

(3)  aS'=  i^>csin(180°-^). 

82 


AREA   OF   AN   OBLIQUE   TRIANGLE  83 

Thus^  the  area  of  a  triangle  is  equal  to 

1-  h)i  sin  A     or     J  he  sin  (180°  -  A) 

according  as  A  is\an  acute  or  an  obtuse  angle. 

While  we  have  found  a  complete  solution  of  our  problem, 

the  result  is  not  quite  as  convenient  as  it  might  be,  since  we 

have  two  different  formulae  for  aS'  alccording  as  A  is  an  acute 

or  an  obtuse  angle.     Is  there  any  way  in  which  we  might 

dvoid  the  distinction  between  these  two  cases  ? 

The  expression  i  ,      .      ^ 

^  be  sin  A 

is  meaningless,  from  our  present  point  of  view,  if  A  is  an 
obtuse  angle.  For  we  have,  as  yet,  given  no  definition  for 
the  sine  of  an  obtuse  angle,  the  definitions  of  Art.  7  being 
applicable  to  acute  angles  only.  Clearly,  however,  it  will 
be  desirable  to  attach  a  meaning  to  the  symbol  sin  A,  also  in 
the  case  when  A  is  an  obtuse  angle,  now  that  we  are  dealing 
with  oblique  triangles,  some  of  whose  angles  may  be  obtuse. 

We  may  define  the  sine  of  an  obtuse  angle  in  any  way  we 
choose,  so  long  as  it  is  not  inconsistent  with  the  definitions  al- 
ready agreed  upon,  and  we  naturally  choose  our  definitions 
and  notations  in  such  a  way  as  to  reduce  to  a  minimum  the 
number  of  formulae  and  theorems  which  must  be  remembered. 

Now  we  can  make  a  single  formula  do  the  work  of  both 
(2)  and  (3),  by  adopting  the  following  definition  for  the  sine 
of  an  obtuse  angle. 

T^e  sine  of  an  obtuse  angle  A  is  equal  to  the  sine  of  the  acute 
angle  180°  —  A,  which  is  supplementary  to  A;  or  in  symbols, 

(4)  sin  A  =  sin  (180°  -  A),  A  being  obtuse. 

As  a  consequence  of  this  definition,  equation  (3),  which 

gives  the  expression  for  the  area  of  the  triangle  when  A  is 

obtuse,  reduces  to  r<      -,  ,      .      . 

JS=:f  be  sm  A, 

so  that  formula  (2)  may  be  used  whether  A  be  acute  or 
obtuse.     The  same  formula  is  obviously  true  when  A  is  a. 


84  THEORY  OF   OBLIQUE   TRIANGLES 

right  angle ;    for  in  that  case  sin  ^  =  1,  and  the  formula  re- 
duces to  S^lhc 

where  c  is  the  base  and  h  the  altitude. 

We  therefore  have  a  single  formula 

S  =^hcmiA 

for  the  area  of  a  triangle  in  terms  of  two  sides  and  the  included 
angle^  whether  the  latter  he  acute,  right,  or  obtuse. 

41.  The  law  of  sines.  Since  the  triangle  has  three  angles 
and  three  pairs  of  including  sides,  we  may  write  three  differ- 
ent expressions  for  the  area  of  the  same  triangle,  viz.  •- 

S=^  ho  sin  A  =  ^  ca  sin  B  —  ^  ah  sin  C. 

The  equality  of  these  three  expressions  is  a  very  importajit 
fact,  since  it  gives  rise  to  the  following  relations  between  the 
sides  and  angles  of  any  triangle : 

he  sin  A  =  ca  sin  ^  =  ah  sin  C. 

We  may  write  these  relations  in  a  somewhat  simpler  form, 
by  dividing  all  three  members  of  the  continued  equation  by 
the  product  ahe.     We  find  in  this  way 

sin  A  _smB  _  sin  O 
a  h  c     ' 

or 

/'I  \  CT     _     ft     _     c 

^^)  8in^~siiiS     sinC' 

whence 

(2)        ■ 


a     sin  A     a     sin  A     h      sinB 


; , : --^ 


h     sin  B     c      sin  O     c      sin  Q 
These  formulae  contain  the  so-called  law  of  sines,  which 
may  be  expressed  in  words  as  follows :  any  two  sides  of  a 
triangle  are  to  each  other  as  the  sines  of  the  opposite  angles. 

The  first  explicit  statement  and  proof  of  the  law  of  sines,  known  at 
the  present  day,  is  to  be  found  in  a  treatise  on  trigonometry  by  the 
Persian,  NasTr  Addin,  or  Nasir  Eddin  (1201-1247  a.d.).  Nasir  Addln's 
treatise  may  also  be  regarded  as  the  first  in  which  trigonometry-  was 
treated  as  a  separate  science,  independent  of  its  applications  to  astron- 
omy. 


THE  LAW  OF  COSINES  85 


EXERCISE  XVIII 


1.  What  becomes  of  the  law  of  sines  when  one  of  the  angles  (say  C) 
is  a  right  angle  ? 

2.  Prove  the  law  of  sines  directly  from  Figs.  34,  35,  36,  by  comput- 
ing the  value  of  h  in  each  of  the  two  right  triangles  into  which  ABC  is 
divided  by  the  altitude. 

3.  Show  that  the  law  of  sines  may  be  used  to  solve  the  following 
problem :  Given  two  angles  of  a  triangle  and  one  of  its  sides ;  to  find 
the  other  sides  and  the  remaining  angle. 

4.  The  formula,  gin  (180°  -  A)  =  sin  A, 

holds  when  A  is  an  obtuse  angle,  as  a  consequence  of  the  definition 
adopted  for  the  sine  of  an  obtuse  angle.  Show  that  the  same  equation 
is  also  true  if  A  is  an  acute  or  right  angle. 

42.  The  law  of  cosines.  A  generalization  of  the  theo- 
rem of  Pythagoras.  We  have  learned  to  recognize  the  im- 
portance of  the  theorem  of  Pythagoras  in  the  theory  of 
right  triangles,  and  the  question  naturally  arises  :  what 
takes  the  place  of  this  theorem  in  the  case  of  an  oblique 
triangle  ? 

Most  students  will  remember  that  the  answer  to  this  ques- 
tion is  contained  in  the  following  two  propositions  of  geom- 
etry :    . 

Theorem  1.  In  any  triangle  the  square  of  the  side  oppo- 
site an  acute  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides  diminished  by  twice  the  product  of  one 
of  those  sides  and  the  projection  of  the  other  upon  that 
side. 

Theorem  2.  In  any  obtuse  triangle,  the  square  of  the 
side  opposite  the  obtuse  angle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  increased  by  twice  the  product 
of  one  of  those  sides  and  the  projection  of  the  other  upon 
that  side. 

The  proof  of  Theorem  1  (repeated  from  Geometry)  is  as 
follows  :  Let  A  be  an  acute  angle.  The  triangle  ABO 
will  have  the  form  represented  in  Figs.  38  or  39,  accord- 


86  THEORY  OF  OBLIQUE   TRIANGLES 

ing  as  the  angle  B  is  acute  or  obtuse.     In  either  case  we 

P^*  BQ=a,     QA^h,     AB^c, 

and  AD  =  m, 

Q    so  that  m  is  the  projection  of  h 
^^x-^y  i      upon    c,   or  upon   c  produced. 
^^c   Ai-c\      The  right  triangle  BOB  gives 


^      m    Dc-mB  A  B  

Fig.  38  Fig.  39  ^2  _  ^2  _j_  ^jf' 

in  both  cases.     Now  BB  =  c  —  m  in  Fig.  38,  and  BB  =  m  —  c 
in  Fig.  39.     Therefore  we  find,  in  either  case, 
a2  =  ^2  +  c2  _  2  (?m  +  m^. 
The  right  triangle  AQB  gives 

in   both  cases.       If   this  value  of   W'  be  substituted  in  the 
equation  above,  we  find 

(1)  «2  __  J2  _|.  ^  _  2  ^/7j  (^A  being  an  acute  angle}^ 

which  proves  Theorem  1. 

To  prove  Theorem  2,  we  refer  to  Fig.  40.     We  have,  in 
this  case, 

a^  =  A2  -f-  ^&  =  A2  +  (6?  4-  w)2 
or  a2=A2  +  c2  +  2cm  +  m2, 


and  Z2        12  2  D  mAv       B 

h^  =  h^—m^,  Fig.  40 

whence 

(2)  a^  =  P -{-  c^  -\-  2  cm  {A  being  an  obtuse  angle), 

which  proves  Theorem  2. 

From  either  Fig.   38  or  39  we  obtain,  by  observing  the 
right  triangle  A  OB, 

m  =  AB  =  b  cos  A, 

In  Fig.  40  we  have  instead, 

^:  CAB  =  180°  -A,   m  =  b  cos  CAB  =  b  cos  (180°  -  A). 
If  we  substitute  these  values  in  (1)  and  (2),  we  find 

(3)  a2  =  /)2  _|_  ^  _  2  l)c  cos  A  (if  A  is  acute), 

(4)  ^2  =  52  +  ^  _^  2  be  cos  (180°  -  A^  (if  A  is  obtuse). 


THE   LAWS   OF   COSINES  87 

m    Just  as  in  Art.  41,  we  have  found  two  different  theorems 
'and  two  different  formulse  for  the  two  cases  when  A  is  an 
acute  or  an  obtuse  angle.     Can  we  again  find  a  single  theorem 
and  a  single  formula  to  do  the  work  of  both  ? 

Equations  (3)  and  (4)  show  that  this  may  indeed  be  done, 
provided  that  we  define  the  cosine  of  an  obtuse  angle  to  he  a 
negative  number^  numerically  equal  to  the  cosine  of  its  supple^ 
ment  (which  is  of  course  an  acute  angle).  For,  with  this 
definition,  we  shall  have 

(5)      cos  A--  cos  (180°  —  A)  {if  A  is  an  obtuse  angle), 

so  that  (4),  as  a  consequence  of  (5),  assumes  the  same  form 
as  (3).  But  formula  (3)  holds  also  when  ^  is  a  right  angle, 
for  in  that  case  cos  ^  =  0,  and  the  formula  reduces  to  the 
theorem  of  Pythagoras.     Thus,  one  and  the  same  formula  (3) 

a^  =  b^  +  c^-2bc  cos  A 

will  he  applicable  to  all  cases  if  it  be  understood^  in  accordance 
with  our  definitions^  that  cos  A  is  positive^  zero^  or  negative 
according  as  the  angle  A  is  acute^  rights  or  obtuse. 

Equation  (3)  is  generally  known  as  the  law  of  cosines,  and 
completely  replaces  Theorems  1  and  2  of  this  Article.  The 
law  of  cosines  obviously  enables  us  to  compute  the  third  side 
of  an  oblique  triangle  when  two  sides  and  the  included 
angle  are  given.  But  it  also  enables  us  to  find  the  angles  of 
a  triangle  when  its  three  sides  are  given.  For,  we  find  from 
(3),  by  transposition, 

2hccosA=b^+(^~a^ 
and  therefore 

The  two  geometric  theorems  (Theorems  1  and  2  of  this  article)  to 
which  the  law  of  cosines  is  equivalent,  were  well  known  to  the  Ancients ; 
and  the  problem  of  finding  the  angles  of  a  triangle,  when  its  three  sides 
are  given,  was  solved  by  Ptolemy  (2d  century  a.d.)  of  Alexandria  in 
his  Almagest  by  means  of  these  theorems.  The  explicit  formulation  of 
the  law  of  cosines,  however,  seems  to  be  due  to  the  great  French  mathe- 
matician, Francois  Viete  (also  known  as  Vieta),  (1540-1603). 


88  THEORY  OF  OBLIQUE   TRIANGLES 


EXERCISE   XIX 

Solve  the  following  triangles,  using  the  tables  of  squares  and  natural 
functions : 

1.  a  =  2,     b  =  3,      C  =  30°.  3.   c  =  2.34,  a  =  4.31,  B  =  116°. 

2.  b  =  3.5,  c  =  2.4,  A  =  52°.  4.   a  =  3,       6  =  6,        c  =  S. 

5.   a  =  1.0,     h  =  2.0,     c  =  1.5. 

^  6.  The  relation  cos^  =  —  cos  (180°  —  ^)  is  true  for  all  obtuse  an- 
gles^ as  a  consequence  of  the  definition  of  the  cosine  of  an  obtuse  angle. 
Prove  that  this  formula  is  also  true  if  A  is  any  acute  angle  or  a  right 
angle. 

•^  7.  Show  that  the  relation  sin^  A  +  cos^  A  =1  holds  for  obtuse  as  well 
as  for  acute  angles. 

8.   If  A  is  an  acute  angle, 

tan  A  =  ?H^,  cot^  =  ^?^,  sec^  =  -^— ,  esc  A  = -'^ 


cos  A  sin^  cos  J.  sin  J. 

Let  us  define  tan  A,  cot  A,  sec  A,  esc  A  by  means  of  these  same  equa- 
tions when  A  is  an  obtuse  angle.  Show  that,  as  a  consequence  of  these 
definitions,  we  have 

tan  ^  =  -  tan  (180°  -A),     cot  ^  =  -  cot  (180°  -A), 
secA  =  -  sec  (180°  -  A),     esc  A  =  esc  (180°  -  A), 
for  any  obtuse  angle  A. 

"      9.   Show  that  the  equations  of  Ex.  8  are  .valid  also  if  the  angle  A  is 
acute. 

10.  The  law  of  cosines  gives  three  equations  for  any  triangle. 
Equation  (3)  of  Art.  42  is  one  of  these.     Write  the  other  two. 

11.  Write  the  equations  for  cos^,  cos  B,  and  cos  C  in  terms  of  the 
tfl^  sides  of  the  triangle. 

12.  Show  that  in  any  triangle, 

a2  +  &2  +  c2  -  2  6c  cos^  -  2  ca  cosB  -  2  ab  cos  C  =  0. 

43.   Properties  of  the  functions  of  an  obtuse  angle.*    We 

have  defined  in  Art.  40  the  sine,  in  Art.  42,  the  cosine,  and 
in  Ex.  8,  Exercise  XIX,  the  remaining  functions  of  an  ob- 
tuse angle.      As  a  consequence  of  these  definitions  we  have 

the  following  system  of  equations  : 

(-, 

*Some  iQstructors  may  prefer  to  change  the  order  of  topics  by  passing  to  the 
discussion  of  the  general  angle,  Art.  60  et  seq.,  and  returning  later  to  Art.  44. 
There  is  no  reason  why  this  should  not  be  done. 


FUNCTIONS  OF  AN  OBTUSE  ANGLE        89 

§rsin  (180°  -A)  =  sin^,         cos  (180° -Jl)  =  -cos  .4, 
)       tan  (180°  -  ^)  =  -  tan  J.,    cot  (180°  -  ^)  =  -  cot^, 
.sec  (180°  -A)=  -  sec  A,    esc  (180°  -  A)  =  esc  A, 

which  are  valid  whether  the  angle  A  be  acute,  right,  or  ob- 
tuse. (Cf.  Exercise  XVIII,  Ex.  4,  Exercise  XIX,  Exs.  6, 
8,  and  9.) 

But  an  obtuse  angle  B  may  be  written  either  in  the  form 

B  =  180° -^,  or ^=  90°  +  A\ 

both  A  and  A'  being  acute  angles.     Moreover,  from 

^=180°-^  =  90°+A' 

follows 

A'  =  90°  -  ^,  or  ^  =  90°  -  A', 

that  is,  the  angles  A  and  A^  are  complementary.  Let  us 
substitute  J.  =  90°  -  A'  in  (1).  We  find  that  the  left  mem- 
ber of  the  first  equation  of  (1)  becomes 

sin  (180°  -A)  =  sin  [180°  -  (90°  -A')]=  sin  (90°  +  A'). 

The  right  member  of  the  same  equation  assumes  the  form 

sin  A  =  sin  (90°  -  ^0  =  cos  J.^         (Art.  10.) 

Consequently,  this  first  equation  of  system  (1)  becomes 

sin  (90°  + J.0  =cos^^    ^ 

In  the  same  way  we  may  prove  the  remaining  equations  of 
the  following  system: 

(sin  (90°  +  A'^  =  cos  J-',       cos  (90°  +  J.')  =  -sin  J.', 
(2)    J  tan  (90°  +^')  =  -  cot^^  cot  (90°  -{-A')  = -ta.nA\ 
I  sec  (90°  +  ^')=  -cscJ.^  csc(90°  + J.O=secJ.'. 

The  student  should  carry  out  the  details  of  these  substi- 
tutions, and  note  the  close  resemblance  between  these  for- 
mulae and  the  formulae  of  Art.  10,  for  sin  (90°  —  A), 
cos  (90°-^),  etc. 

This  resemblance  is  so  close  as  to  make  it  easy  to  remember  equations 
(2).  Their  right  members  differ  from  the  right  members  of  the  corre- 
sponding equations  for  sin  (90°  —  A),  cos  (90^^  -A},  etc.,  at  most  in 
sign.     This  remark  suffices  to  help  us  remember    that  sin   (90°  +  A) 


90  THEORY  OF   OBLIQUE   TRIANGLES 

is  either  equal  to  +  cos  A  or  to  —  cos^,  that  cos  (90°  -\-  A)  is  eithei 
equal  to  +  sin  A  or  —  sin  ^,  etc.  In  order  to  choose  between  these 
alternatives  we  may  argue  as  follows.  Let  A  be  an  acute  angle.  Then 
sin  A  and  cos  A  are  both  positive.  But  since  90°  +  A  will  then  be  ob- 
tuse, sin  (90°  +  ^)  is  positive  and  cos  (90  +  ^)  is  negative.  Therefore, 
of  the  two  alternatives 

sin  (90°  +  A)  =  cos  A,  or  sin  (90°  +  A)  =  -cos  A, 

we  must  discard  the  latter,  since  its  left  member  is  positive  and  its  right 
member  is  negative.     Similarly,  of  the  two  alternatives 

cos  (90°  +  A)  =  sin  A,  or  cos  (90  +  A)  =  -  sin  A, 

we  must  discard  the  former.  For  it  involves  the  contradiction  that  the 
negative  number  cos  (90°  +  A)  should  be  equal  to  the  positive  number 
sin  A.     This  argument  fixes  the  two  formulae 

sin  (90°  +  A)  =  cos  A,  cos  (90°  +  ^)  =  -  sin  ^ 

in  our  memory.  The  remaining  formulae  of  system  (2)  may  be  remem- 
bered in  similar  fashion. 

The  first  two  formulae  of  system  (1),  namely, 

sin  (180°  -  A)  =  sin  A,  cos  (180°  -  A)  =  -  cos  A, 

are  easy  to  remember,  since  it  was  upon  these  equations  that  we  based 
our  definitions  of  the  sine  and  cosine  of  an  obtuse  angle.  The  remain- 
ing formulae  of  system  (1)  follow  directly  from  these  two. 

EXERCISE  XX 

1.  Express  the  following  quantities  as  functions  of  acute  angles; 
sin  100°,  cos  115°,  tan  162°,  cot  99°,  sec  120°,  esc  175°. 

2.  By  means  of  the  table  of  natural  functions,  find  sin  98°.5, 
cos  176°.3,  tan  124°.7,  cot  134°.6. 

3.  Explain  how  equations  (1)  and  (2)  of  Art.  43  provide  two  differ- 
ent methods  of  finding  the  functions  of  obtuse  angles  from  the  table  for 
acute  angles. 

44.   Other  formulae  for  the  area  of  an  oblique  triangle.     The 

methods  of  Art.  40  suffice  to  determine  the  area  of  a  tri- 
angle if  one  side  and  the  corresponding  altitude  (<?  and  A), 
or  if  two  sides  and  the  included  angle  (6,  e,  and  A)  are 
given. 

Let  us  suppose  now  that  one  side  and  two  adjacent  angles 


ADDITIONAL  AREA  FORMULA  91 

((?,  A,  B)   are    given.     We  have  in   the   first    place    from 

Art.  40 

(1)  S=^hc^\uA, 

and  it  only  remains  to  modify  this  formula  by  expressing  h 
in  terms  of  c,  A^  and  B.     The  law  of  sines,  in  the  form 

ro\  ^  —  si"  B 

and  the  equation 

(3)  A-hB-h  0=  180° 

enable  us  to  do  this.     For  we  find  from  (2)  and  (3) 
c  sin  B  c  sin  B  c  sin  B 


b  = 


sin  0      sin  [180°  -  (A  +  B)]      sin  (A  +  B) 

since,  according  to  Art.  43,  the  sine  of  any  acute  or  obtuse 
angle  is  equal  to  the  sine  of  its  supplement. 

If  we  substitute  this  value  of  h  in  (1),  we  find 

^4-.  a  _   (^  sin  A  sin  B 

^  ^  ~2sin  (A  +  By 

the  desired  formula  for  S  in  terms  of  c,  A,  and  B. 

If  one  side  c  and  two  angles,  not  both  adjacent  to  e,  are 
given,  we  may  first  find  the  third  angle  from  (3)  and  then 
use  formula  (4). 

Let  us,  finally,  obtain  a  formula  for  S  in  terms  of  the 
three  sides  a,  6,  c.  We  start  again  from  equation  (1)  which 
already  contains  the  sides  h  and  c  of  the  triangle,  and  which 
will  give  the  desired  formula  if  we  can  express  sin^  in 
terms  of  a,  5,  and  c.  This  may  be  done  by  making  use  of 
the  law  of  cosines  (which  gives  cos  A  in  terms  of  a,  6,  and 
<?),  and  the  relation 

(5)  sin2^  +  cos2^  =  1, 

which  holds  for  obtuse  as  well  as  for  acute  angles.     (See 
Ex.  7,  Exercise  XIX.) 

To  avoid  the  inconvenience  of  writing  cumbersome  square 


92  THEORY  OF   OBLIQUE   TRIANGLES 

root  signs,  let  us  square  both  members  of  (1).  We  find, 
making  use  of  (5), 

•      ,S^2  =  1  J2^  sin2  ^  =  1  6V  (1  -  cos2  A} 

or,  factoring  the  binomial  on  the  right, 

^  =  1  JV  (1  +  cos ^)  (1  -  cos ^). 

By  the  law  of  cosines  (Equation  (6),  Art.  42),  this  becomes 

^h^     2  ^g  +  52  +  g2  -  gg     2bc-b^-c^  +  a^ 
4    *  2  he  '  2  be 

Each  of  the  factors  on  the  right  member  may  be  factored, 
giving  the  equation 

(6)  S^  =  ^Q  (J  +  c  +  g)  (5  +  c-«)  (^a+b-e)  (a-5  +  c?). 

The  first  factor  on  the  right  member  of  (6)  is  the  perimeter 
of  the  triangle,  and  the  formula  becomes  especially  simple 
if  we  denote  half  of  this  perimeter  by  s,  so  that 

(7)  a  +  5  4-  <?  =  2  s. 

The  other  three  factors  of  the  right  member  of  (6)  will  then 
become 

—  a-\-b  +  e=2s  —  2a  =  2  (s  —  a), 

(8)  a-b  +  e=28-2b  =  2  (8-b), 
a-\-b-c=28-2e=2(8~e}, 

so  that  (6)  reduces  to 

^2,,,^  (s-a)  («-J)  («-c), 

whence  finally,  since  aS'  is  positive, 

(9)  8  =  Vs  (8 -a)  (8-b)  («-c), 

a  famous  equation  generally  known  as  Hero's  formula,  after 
Hero  of  Alexandria,  who  lived  about  120  B.C.,  and  wrote 
a  famous  textbook  on  surveying. 


RADIUS  OF  INSCRIBED  CIRCLE 


93 


EXERCISE  XXI 
Find  the  area  of  the  following  triangles  to  four  significant  figures. 

1.  a  =  2,     ?>  =  3,      C  =  30°.  4.   a  =  3,    A=  30°,  B  =  75°. 

2.  b  =  3.5,  c  =  2.4,  A  =  52°.  5.   a  =  3,     6  =  6,       c  =  8. 

3.  c  =  5,  A  =  30°,  5  =  75°.  6.   a  =  1.0,  6  =  2.0,    c  =  1.5. 

45.    The  radius  and  center  of  the  inscribed  circle.     The 

fundamental  basis  for  all  of  the  different  formulse  which  we 
have  obtained  for  the  area  of  a  triangle,  so  far,  was  the 
equation  S=lch, 

from  which  all  of  the  others  were  derived. 

But  this  formula  is  unsymmetrical,  since  it  singles  out  one 
of  the  sides  of  the  triangle  and  subjects  it  to  a  treatment 
different  from  that  accorded  to  the 
other  two.  We  may  avoid  this  lack 
of  symmetry  by  picking  out  a  point 
M  anywhere  inside  of  the  triangle 
ABO  (see  Fig.  41),  and  joining  M 
to  the  three  vertices.  The  area  of 
the  given  triangle  will  then  appear 
as  the  sum  of  the  areas  of  the  three 

triangles  BMC,  CMA,  AMB,  whose  altitudes  are  respec- 
tively equal  to  MD,  ME,  and  MF. 

Clearly,  there  is  one  position  of  the  point  iltf  which  is  better 
adapted  for  this  purpose  than  any  other,  namely  the  center 
0  of  the  inscribed  circle.  For  the  distances  from  0  to  the 
three  sides  of  the  triangle  are  equal  to  each  other,  so  that 
the  three  triangles  BOO,   00 A,  and  AOB  will  have  equal 

altitudes. 

Let  0  (Fig.  42)  be  the  center  of 
the  inscribed  circle  and  let  r  be  the 
radius  of  this  circle. 

Then  the  area  of  the  triangle  is 
S=B00^-  OOA+AOB 
=z  ^  ar  +  ^  br  -{-  \  cr 

Fig.  42  =  J  («  +  6  +  c)  T, 


Fig.  41 


94  THEORY   OF  OBLIQUE   TRIANGLES 


or  finally 

(1)  S=sr,    " 

where  s  denotes  the   half -perimeter   of  the  triangle   as   in 
Art.  44. 

Since  we  also  have 


S  =.^ %{8-a){^-h){s-c)    (Equation  (9),  Art.  44), 

we  find  from  (1),  substituting  this  value  of  S  and  dividing 
both  riiembers  of  the  resulting  equation  by  «, 

(2)  ^^^{s-a){s-b)(8-c)^ 

which  enables  us  to  compute  the  radius  of  the  inscribed  circle 
when  the  sides  of  the  triangle  are  given. 

If  we  wish  to  know,  not  merely  the  radius  of  the  inscribed 
circle,  but  also  the  position  of  its  center,  we  must  compute 
the  lengths  of  the  line  segments  AF^  BF,  etc.  in  Fig.  42. 
Now  we  know  from  Geometry  that 

(3)  AF  =  AU,  BB  =  BF,  CE  =  CD, 

Moreover,  the  sum  of  these  six  line-segments  is  equal  to  the 
perimeter  2s  of  the  triangle.  Therefore  the  sum  of  three  of 
these  segments,  one  chosen  from  each  of  the  three  equations 
(3),  will  be  equal  to  half  of  the  perimeter.     That  is, 

AF-i-  BB-^  OB  =  8, 
But 

BB+  CB  =  a, 

and  therefore 

(4)  AF=AE=8-a, 
In  the  same  way  we  find 

(5)  (BB  =  BF=8-b, 
1  CF=  0B  =  8-c. 

The  three  equations,  (4)  and  (5),  enable  us  to  locate  the 
points  D,  F,  F  in  which  the  inscribed  circle  touches  the  sides 
of  the  triangle  and  consequently  determine  completely  the 
position  of  the  center  0  of  the  circle. 


RADIUS  OF   CIRCUMSCRIBED   CIRCLE  95 

46.  The  half -angle  formulae.  Since  the  center  0  of  the 
inscribed  circle  is  the  point  of  intersection  of  the  three 
angle  bisectors  of  the  triangle  (see  Fig.  42),  the  angle 
FAO  is  equal  to  |  A.  In  the  right  triangle  FAO^  we  have 
therefore 

But  FO  is  the  radius  r  of  the  inscribed  circle,  and  AF  we 
have  just  found  to  be  equal  to  8  —  a.  (Equation  (4),  Art. 
45.)     We  find  therefore 

and  in  the  same  way 

r 
.  laii  -  iv  =  — 

8-b 


(2)  tan|l^  =  -f-,tan|c  =  — -, 


the  value  of  r  being  given  by  equation  (2)  of  Art.  45. 

These  three  equations,  usually  known  as  the  half-angle 
formulae,  are  very  important  in  providing  a  second  method 
for  computing  the  angles  of  a  triangle  when  its  sides  are 
given.  They  are  far  more  convenient  for  this  purpose  than 
the  law  of  cosines,  if  logarithms  are  to  be  used.  In  fact,  the 
law  of  cosines  is  so  cumbersome  from  the  point  of  view  of 
logarithmic  calculation,  that  we  shall  seek  to  find  a  substi- 
tute for  it  also  in  the  only  other  case  in  which  we  have  pro- 
posed to  make  any  use  of  it,  namely  in  the  solution  of  a 
triangle  when  two  sides  and  the  included  angle  are  given. 
(See  Art.  42,  p.  96.) 

We  should  not  neglect  to  note,  however,  that  the  law  of  cosines  has 
in  recent  times  again  come  into  practical  use,  especially  in  engineering 
practice,  the  calculations  being  performed  not  with  logarithms,  but  with 
the  help  of  tables  of  squares  and  products  or  with  a  calculating  machine. 
The  Eichhorn  Trigonometric  slide  rule,  mentioned  on  page  66,  is  based 
entirely  upon  the  law  of  cosines. 

47.  The  circumscribed  circle.  The  center  0  of  the  circum- 
scribed circle  is  the  point  of  intersection  of  the  three  per- 
pendicular bisectors  of  the  sides  of  the  triangle.     Therefore, 


96 


THEORY  OF  OBLIQUE   TRIANGLES 


if  iVis  the  middle  point  of  the  side  AB  of  the  triangle  (see 
Fig.  43),  the  line  OiV^will  be  perpendicular  to  AB. 

Now  angle  AOB  is  measured  by  the  arc  AB^  and  the 
angle   AOB  is   measured   by   half 
this  arc  (being  an  inscribed  angle). 
Therefore 

ZAON=ZAOB=^  O 
since  each  of  these  angles  is  equal 
Ib  to  half  of  z  AOB.     Consequently 
we  find,  making  use  of  the   right 
triangle  A  OiV", 


Fig.  43 


sin  0=  sin A0N=4^, 
AO 


If  we  denote  AO,  the  radius  of  the  circumscribed  circle,  by 
B,  this  becomes 


sm  (7=2^  =  -—— 
B     2B 


,or2i2  = 


sin  0 
whence,  making  use  of  the  law  of  sines  (Equation  (1),  Art. 

41), 

a  b  c 


(1) 


2B  = 


sin  A     sin  B     sin  O'' 
thus  completing  the  law  of  sines  by  providing  a  simple  geo- 
metrical interpretation  for  the  common  value  of  the  three 
equal  ratios  a  h  c 

sin^'    sin^'    sin  O^ 
namely,  the  diameter  of  the  circumscribed  circle. 
Since  the  area  of  the  triangle  is 

S=  ^hc  sin  A, 
and  since  we  find,  from  (1), 

sm  A  =  — — , 
2B 

we  obtain  the  following  remarkable  formula 

abc 


(2) 

for  the  area  of  the  triangle. 


JS  = 


4.B 


THE   FORM  RATIOS  OF  A   TRIANGLE  97 

EXERCISE  XXII 

1.  Compute  the  radius  of  the  inscribed  circJe  and  the  lengths  of  the 
line-segments  into  which  this  circle  divides  the  sides  of  the  triangle 
whose  sides  are  a  =  3,  6  =  6,  c  =  8. 

2.  What  relation  will  there  be  between  the  sides  of  a  triangle  if  the 
inscribed  circle  bisects  one  of  its  sides  ?  If  it  touches  one  of  the  sides  at 
a  trisection  point  ? 

3.  Find  formulae  for  the  distances  from  the  center  of  the  inscribed 
circle  to  the  three  vertices  of  the  triangle. 

4.  Making  use  of  the  results  of  Ex.  3,  show  that 


sin  ^  ^  =  V  (s  —  &)(s  -  c)lhc  and  cos  ^  ^  =  v.s(a-  —  a)lhc. 

5.  Find  a  formula  for  the  area  of  the  inscribed  circle,  and  for  the 
ratio  of  this  area  to  that  of  the  triangle  itself. 

6.  By  means  of  the  formulae  of  Ex.  5,  show  that 


"^^         ^r  <^^ 

7.  Show  that 

r  =  s  tan  |  A  tan  ^  B  tan  ^  C. 

8.  A  circle  is  said  to  be  escribed  to  a  triangle,  or  inscribed  externally, 
if  it  touches  one  of  the  sides  externally  and  the  prolongations  of  the  other 
two  sides  (Fig.  44).     There  are  three  such 

circles    for   every  triangle.     Let  r^  be  the  m^^ 

radius  of  that  one  which  touches  the  side  a 
externally.     Show  that 

AM=AN=s  

and  A  B~ 

r,  =  s  t an  i  ^  =  -^  =  ^:^ .  Fig  .  44 

s  —  a      s  —  a 

9.  If  Ta,  rj,  ?'c,  are  the  radii  of  the  three  escribed  circles,  and  r  de- 
notes the  radius  of  the  inscribed  circle,  show  that 

r      ra      n      r^      . 
10.   Show  that 

r-a  +  r-fe  +  re  -  r  =  4  jR, 

if  R  is  the  radius  of  the  circumscribed  circle. 

48.  The  form  ratios  of  a  triangle.  If  two  sides  of  a 
triangle  are  equal,  the  triangle  is  said  to  be  isosceles  ;  if  no 
two  sides  are  equal,  it  is  usually  called  a  scalene  triangle. 
But  if  the  difference  between  two  sides  of  a  triangle  is  small, 


98  THEORY  OF  OBLIQUE   TRIANGLES 

as  compared  with  the  combined  length  of  these  two  sides,  the 
triangle  will  differ  but  little  from  the  isosceles  form.  Con- 
sequently, the  ratio  of  the  difference  between  two  sides  of  a 
triangle  to  their  combined  length  may  be  taken  as  a  nu- 
merical measure  for  the  departure  of  the  triangle  from  the 
isosceles  form,  or  as  its  form  ratio  with  respect  to  those  two 
sides. 

There  are  three  such  form  ratios  for  every  triangle.  If 
we  assume  that  the  notation  be  so  chosen  that 

a>h^c, 

these  three  form  ratios  are 

a  —  b         a—  c         b  —  c 

a+  b         a-{-  c         b  -{-  c 

Clearly,  one  of  them  will  be  equal  to  zero  if  the  triangle  is 
isosceles,  and  all  three  will  vanish  for  an  equilateral  tri- 
angle. 

By  means  of  the  law  of  sines,  each  of  these  form  ratios 
may  be  expressed  in  terms  of  the  angles  of  the  triangle.  In 
fact,  according  to  Art.  47,  equation  (1),  we  have 

a=2EsmA,     b=2BsinB,     c=2BiimO, 

so  that  we  obtain  the  expression 

r^\  a  —  b  _  sin  ^  —  sin  ^ 

^  a  +  b      sin  ^  -f  sin  jB 

for  the  form  ratio  with  respect  to  the  sides  a  and  b.  Simi- 
lar expressions  hold,  of  course,  for  the  other  form  ratios  ; 

^^«  and  l^. 

a  +  c  b  -\-  c  ■    ' 

49.   The  formulae  for  the  sum  and  difference  of  two  sines. 

The  fraction  which  occurs  in  the  right-hand  member  pf 
equation  (1),  Art.  48,  is  not  adapted  to  logarithmic  calcula- 
tion. But  we  shall  show  that  both  numerator  and  denom- 
inator of  this  fraction,  namely,  the  expressions  sin  J.—  sinB, 
and  sin  A  -f  sin  5,  may  be  written  in  the  form  of  products, 
thus  making  it  easy  to  compute  their  values  by  logarithms. 


THE   SUM   AND  DIFFERENCE   OF   TWO   SINES 


99 


To  discover  the  product  form  of  these  expressions,  we 
must  construct  a  figure  in  which  the  angles  A  and  B^  the 
sines  of  these  angles,  and  the  sum 
and  difference  of  their  sines,  shall 
appear. 

We  construct  first  (Fig.  45)  the 
angles 

Z  XOF  =  ^  and  Z  XOQ=  B. 

With  their  common  vertex    0  as 
center,  and  any  convenient  radius 
r,  we  draw  a  circle  whose  intersec- 
tions with  the  sides  of  the  angles  we  denote  by  X,  P,  and  Q, 
respectively. 

Let  PM  and  QN  be  perpendicular,  and   QR  parallel  to 
OX.     Then 

(1)  r  sin  ^  =  MP,  and  r  sin  j&  =  NQ. 

But  MP  and  NQ  are  the  two  bases  of  the  trapezoid  MPQN. 
If  S  is  the  middle  point  of  the  chord  PQ,  and  US  be  drawn 
perpendicular  to  OX,  US  is  the  median  of  this  trapezoid,  and 
we  shall  have, 

US  =  l(iMP-^NQ\ 

whence,  according  to  (1), 

(2)  r  (sin^  +  sin  J5)  =  MP+WQ  =  2  US. 
On  the  other  hand. 


TS=iPB 


I  (MP 


^Q}. 


and  hence, 

(3)  r(sin^  -  sin5)  =  MP  -  NQ  =  2  TS. 

In  order  to  express  the  right  members  of  (2)  and  (3)  in 
terms  of  r,  A  and  B,  we  consider  the  right  triangles  OSU 
and  SQ_T  in  which  US  and  TS  occur.  It  is  apparent  from 
the  figure  that 

USOQ=lQOP=iCA-B}, 

[Z  UOS  =  5  +  I  (  A-  5)  =  1  (^  +5), 


(4) 


100  THEORY  OF   OBLIQUE   TRIANGLES 

and 

(5)  ATSQ^Z.  UOS  =  H^  +  i5), 

since  each  of   these  angles  is  complementary  to  the  angle 
USD. 

Hence  we  have  from  the  right  triangles  OiS'ZJand  SQT, 

US  =  OS&in  UOS  =  OS  sin  ^{A  +  B\ 
^  ^  TS=SQ  cos  TSQ  =  SQ  cos  -i  (^  +  ^), 

and  from  the  triangle  OQS, 

OS  =  OQ  cos  SOQ  =  r  cos  -|  (^  -  ^), 
^  ^  ^§  =  OQ^inSOQ  =  rsin  1  (^  -  ^). 

Substituting  (6)  and  (7)  in  (2)  and  (3)  and  dividing  by 
/•,  we  find 

sin^.  +  sin^  =  2sm^  (^  +  B)  cos^  (^  -JB), 

(8)  1  1 

^  ^  sin^  -  sin  J5  =  2cos|  (^  +  B)  mn\  {A  -  B), 

which  are  the  desired  formulae. 

The  student  should  observe  that  our  proof  of  equations 
(8),  based  on  Fig.  45,  remains  valid  even  if  A  is  an  obtuse 
angle,  as  long  as  MP  is  greater  than  or  equal  to  NQ ;  that  is, 
as  long  as  ^  +  ^  does  not  exceed  180°.  We  may  therefore 
apply  equations  (8)  whenever  A  and  B  are  two  angles  of  the 
same  triangle,  since  the  sum  of  two  angles  of  a  triangle  can 
never  exceed  180°. 

50.  The  law  of  tangents.  Let  us  now  return  to  the  expres- 
sion (1)  of  Art.  48  for  the  form  ratio  of  the  triangle  with 
respBct  to  the  sides  a  and  6,  of  which  sides  we  shall  assume 
a  to  be  the  greater.     We  had  found 

a  —  b  _  sin  ^  —  sin  ^ 

a  -\-  h      sin  A  4-  sin  B 

If  now  we  make  use  of  equations  (8)  of  Art  49,  we  find 

g  -  5  ^  2  cos  jr  C^  4-  ^)  sin  1  (^  -  -^) 
a  +  b      2  sin  1-  (^  +  B)  cos  ^  (A  -  B)' 


THE  LAW  OF   TANaii:XTS    v  101 

whence  -    - 

I  a  +  0 

since  we  have,  for  any  angle  M, 

sin  M     .       Ttyr    cos  M         .   -mm- 

—  =  tan  M^    =  cot  M. 

cos  M  sin  M 

But  the  tangent  and  cotangent  of  the  same  angle  are  re- 
ciprocals, so  that  we  may  write  finally 

a-b     tan  I  (^  -  B) 

a  very  important  formula,  generally  known  as  the  law  of 
tangents.     We  find  in  the  same  way : 


(2) 


a  —  c  _  tan  \  {A  —  O) 

a  +  (?"~tanl  (J.  4-  Oy 

5-g^tanK-g-  O 
h  +  c      tan^  (^  +  oy 


The  law  of  tangents  seems  to  have  been  expressed  in  this  form  for  the 
first  time  by  Vieta,  to  whom  we  also  owe  the  modern  form  of  the  law  of 
cosines.  It  had  however  been  stated,  in  a  more  complicated  but  equiva- 
lent form,  about  ten  years  earlier  by  the  Dutch  mathematician  Thomas 
Fink  or  Finchius  (1561-1656)  in  his  Geometria  rotundi.  It  was  also 
Fink  who  first  introduced  the  names  tangent  and  secant  for  the  functions 
which  we  now  call  by  these  names.  Many  previous  authors  had  used  the 
name  umbra  =  shadow  for  the  function  which  we  now  call  tangent,  on 
account  of  the  relation  of  this  function  to  the  shadow  cast  by  a  vertical 
stick.  (Compare  Art,  3  and  solve  the  problem  there  attributed  to  Thales 
by  trigonometry.) 

Fink  not  only  discovered  the  law  of  tangents,  but  pointed  out  its 
principal  application;  namely,  to  aid  in  solving  a  triangle  when  two 
sides  and  the  included  angle  are  given.  The  possibility  of  such  an  ap- 
plication will  appear  from  the  following 

Illustrative  Example.  Given  a,  b,  and  C.  To  find  A,  B,  and  C. 
Solution.     The  law  of  tangents  (Equation  (1))  gives 

(3)  tan  1  (^  -  B)  =  ^—\  tan  ^(.4  -\-  B), 

a  +  0 


102 


TfiEOftY  OF   OBLIQUE   TRIANGLES 


aiV  equaiiioii  iiv  which  the  right  member  is  completely  known,  since  a,  h, 
and  C  are  given,  and  since 

(4)  \{A  +B)  =  K180°  -C)=  90°-i  a 

Thus  we  can  compute  tan  l(A  —  B)  by  means  of  (3)  and  then  find 
l(A  —  B)  from  the  table.  Hence,  knowing  \(^A  —  B)  and  \{A  -\-  B)  we 
can  find 

(5)  A  =\{A  +B)+l{A  -B), 
B  =  i(A  +B)~l(A  -B). 

The  sine  law  now  enables  us  to  find  c,  since 


(6) 


sin  C 


gives  c  = 


a  sin  C 


sin  A  sin  A 

If  we  wish  to  solve  the  same  problem  by  the  law  of  cosines,  we  first 
compute  c  from  the  equation 
(7)  c2  =  a2  +  &2  _  2ab  cos  C 

and  afterward  determine  the  angles  A  and  5  by  using  the  law  of  sines. 

The  first  method  has  the  advantage  over  the  second  that  formulae 
(3),  (4),  (5),  (6)  are  in  a  convenient  form  for  logarithmic  computatipn, 
while  equation  (7)  is  not. 

51.  A  second  proof  of  the  law  of  tangents  and  Mollweide's 
equations.     The  law  of  tangents  may  also  be  proved  directly 

from  a  figure 
without  making 
use  of  the  for- 
mulae for 

sin  A  +  sin  B 
and 
sin  A  —  sin  B. 

Let  BC  =  a 

and       CA  =  h 
(Fig.     46)     be 
Fig.  46  two  sides  of  the 

triangle  ABO^  and  let  a>h.  Draw  a  circumference  with  C 
as  center  and  radius  equal  to  6,  and  let  I)  and  E  be  the 
points  in  which  this  circumference  meets  BC  and  BO  pro- 
duced. Let  F  be  the  point  of  intersection  of  the  circum- 
ference  with  AB.     Then  we  have 


SECOND   PROOF  OF  LAW  OF  TANGENTS  108 

(1)  BD^a-h,  BE=a  +  h. 

Since  Z.  EC  A  is  an  exterior  angle  of  the  triangle  ABO^  the 
opposite  interior  angles  being  A  and  jB,  we  have 

Z  EGA  =  A  +  B, 

so  that 

(2)  Z  EBA  =  liA  +  B) 

since  the  latter  angle,  being  inscribed  in  the  circumference, 
is  measured  by  half  the  arc  which  it  subtends. 

It  remains  to  find  an  angle  in  Fig.  46,  equal  to  \(^A  —  ^). 
In  order  to  do  this,  let  us  draw  CF.  Since  A  OF  is  an  isos- 
celes triangle,  we  have 

AAFQ=^iFA'C=A 

and,  since  Z  AFC  is  an  exterior  angle  of  the  triangle  BFC^ 

A=  /.  FOB  +  B, 

whence 

Z FOB  =  A-  B. 

Since  Z  DAF  is  an  inscribed  angle  subtending  the  same  arc, 

(3)  Z  BAB  =  Z  BAF  =  i(^  -  ^). 

Let  us  apply  the  law  of  sines  to  the  two  triangles  ABB 
and  ABE,     We  find,  from  the  first  triangle, 

a-h^^inBAB^     sinl(^-^)      ^sinl(J.-^) 
c         sin  ABB     sin  (1S0°  -  EBA}  sin  EBA 

or  finally 

.^.  a-5  ^  sin-K^--^) 

^  ^  (?         sinl(^  +  ^)* 

Similarly  the  triangle  ABE  gives 

a-\-h  ^sin  EAB 
c     ~  sin  AEB' 
But 

Z  ^^5  =  Z  J5;^2)  H-  Z  i)^J5  =  90°  +  J  (^  -  ^)' 
Z  ^J5^^  =  90°  -  Z  JS'i)A  =  90°  -  1  (^  +  B}, 

owing  to  (2)  and  (3)  and  the  further  fact  that  Z  EAB  is  a 
right  angle,  since  it  is  inscribed  in  a  semi-circumference. 


104  THEORY   OF   OBLIQUE   TRIANQLES 

Therefore 

a  +  h^  sin  [90°  +  K^  -  -g)1 
c         sin  [90°  -  K^  +  ^)]' 
which  may  be  written 

^  ^  0    cosi(^+^y 

as  a  consequence  of  equations  (2),  Art.  43,  and  of  Art.  10. 

If  now  we  divide  equations  (4)  and  (5),  member  by  mem- 
ber, we  find 

a-b  ^sin^  (A-  B)    cos  l(A-\-  B)  ^  tan  i  ( J.  -  jg)  . 
a-\-b      sinl(J.  +  5)    gos^Ia-B)      tan  i  (  JL  +  5)  ' 

that  is,  the  law  of  tangents. 

Incidentally  we  have  found  two  new  formulse,  (4)  and 
(5).  They  assume  a  somewhat  more  serviceable  form  by 
means  of  the  relation  A-\-  B  +  (7=  180°,  which  gives 

and  therefore 

sin  J  (A  +  ^)  =  cos  J  O,   cos  |  (^A-\-B)  =  sin  ^  0^. 

If  we  introduce  these  values  in  equations  (4)  and  (5), 
they  become 

a_6     sin^C^-S)     ^^i,     cosliA-B) 


(6) 


cos^C  ^  smlc 


These  formulce^  known  as  the  Mollweide  equations,  are 
particularly  convenient  for  the  purpose  of  checking  the  accuracy 
of  the  numerical  solution  of  a  triangle.  For  each  of  these 
equations  contains  all  of  the  six  parts  of  the  triangle,  so  that 
an  error  in  any  one  of  these  parts  would  be  likely  to  make 
itself  felt  by  a  lack  of  agreement  between  the  two  members 
of  one  of  these  equations. 


THIRD   PROOF   OF  LAW  OF  TANGENTS  105 


Of  course  there  are  two  other  pairs  of  equations  of  the 
rm  (6),  t 
h  —  c    h  -\-  c 


form  (6),  the  left  members  of  which  are  — - — ,      7^    ,  and 


,  respectively. 
a  a 

It  is  not  justifiable  historically  to  call  equations   (6)   MoUweide's 

equations.     The  formula  for  ^  is  to  be  found  in  Newton's  Arith- 

c 

metica  Universalis.  Both  equations  (6)  are  given  in  Simpson's  Trigo- 
nometry, Plane  and  Spherical  (1748),  and  also  in  F.  W.  von  Oppel's 
Analysis  Triangulorum  (1746).  All  of  these  works  antedate  considerably 
the  publication  of  these  equations  by  Mollweide  in  1808. 

EXERCISE  XXIII 

1.  Show  that  the  law  of  tangents  may  also  be  obtained  from 
Fig.  46  by  drawing  through  D  a  line  parallel  to  AE  and  meeting  the 
side  AB  in  G,  and  then  computing  tan|  (A  —  B)  and  tan^  {A  +  J5) 
from  the  right  triangles  ADG  and  A  ED. 

2.  Still  another  proof  of  the  law  of  tangents  proceeds  as  follows. 
In  Fig.  47,  draw  CD,  the  bisector  of  angle  C,  and  drop  perpendiculars, 
AL  and  BM,  from  A  and  B  to  CD. 
Let  N  be  the  point  of  intersection 
ot  AB  with  CD.     Then 

ZBCD  =  ZDCA  =iC 
and 
(1)        ZLAN=ZNBM 

=z90°  -Z  BNM.  ^^^'  *^ 

But  Z  BNM  is  an  exterior  angle  of  the  triangle  BCN,  so  that 

Z  BNM  =B-hiC. 
Moreover, 

90'^  =  l(A  -{-B  +  C) 

and  therefore,  by  substitution  in  (1), 

ZLAN  =  i(A  +  £  +  C)  -  (5  +  i  C)  =  K^  -  B). 

Now,  from  the  right  triangles  ANL  and  BNM,  we  see  that 

4.       ^,A        Dx      LN     MN 
tani(^-5)=  — =— , 

and  therefore  also 

LN  +  MN 


(2)  tan  i(A  -  B) 


LA  -^MB 


106  THEORY  OF   OBLIQUE    TRIANGLES 

But 

LN  +  MN  =  CM  -  CL  =  a  cos^C-b  cos  ^  Cz=  (a  -  b)cos ^  C, 

LA  +  MB  =  6  sin  i  C  +  a  sin  ^  C  =  (a  +  J)  sin  ^  C, 
which  gives,  on  substitution  in  (2), 

tan  i(A  -B)  =  -^=-^cot  ^  C, 
a  -\-h 

and  this  is  equivalent  to  the  law  of  tangents,  since 
C  =  180°  -(A  +  B). 


CHAPTER   VIII 

SOLUTION  or  OBLIQUE  TRIANGLES 

52.  The  fundamental  problem.  Each  of  the  laws  found  in 
Chapter  VII  contains  four  of  the  six  parts  of  a  triangle,  and 
thus  suggests  the  possibility  of  computing  any  one  of  these 
four  parts  when  the  other  three  parts  which  occur  in  that 
law  are  given. 

On  the  other  hand,  the  relation 
(1)  A-^B+  0=  180°, 

which  is  true  for  every  triangle,  contains  only  three  of  its 
parts.  The  three  angles  of  a  triangle  are  therefore  not  inde- 
pendent of  each  other,  as  is  the  case  with  the  three  sides. 
The  familiar  fact,  that  a  triangle  is  not  determined  by  its 
three  angles,  is  to  be  regarded  as  a  consequence  of  this. 
For,  on  account  of  relation  (1),  the  three  angles  of  a 
triangle  are  not  independent  data. 

Now  there  exists  no  other  equation  which^  like  (1),  contains  no 
more  than  three  parts  of  the  triangle  and  which  is  true  for  all 
triangles.  For,  if  there  were  such  an  equation,  for  instance, 
between  a,  5,  and  c,  it  would  be  impossible  to  find  a  triangle 
in  which  more  than  two  of  these  quantities  have  arbitrarily 
assigned  values  since  two  of  the  three  quantities  would  then 
determine  the  third.  But  this  is  contrary  to  well-known 
facts  of  geometry ;  since  a  triangle  may  be  constructed  for 
which  all  three  sides  a,  5,  and  c  have  arbitrarily  assigned 
values,  provided  only  that  a  ■\-h>  c.  The  values  of  a  and  h 
do  not  determine  the  value  of  c ;  while  the  values  of  A  and  B 
do  determine  the  value  of  C. 


k 


Our  illustration  shows  incideutally  that  there  may,  and,  in  fact,  do 
exist,  besides  the  equation  (1),  certain  inequalities  between  three  parts  of 

triangle  which   must  be  satisfied  by  all  triangles.     The  inequality 
a  +  ft  >  cis  an  instance. 

107 


108  SOLUTION   OF   OBLIQUE   TRIANGLES 

Any  three  parts  of  a  triangle,  unless  all  three  of  them  are 
angles,  may  therefore  be  regarded  as  independent  data. 
Consequently  there  arises  the  following  fundamental  problem : 

To  find  the  remaining  parts  of  a  triangle  when  any  three  in- 
dependent parts  are  given. 

The  discussion  of  this  problem  leads  to  a  division  into  four, 
cases. 

Case  I.    One  side  and  two  angles  are  given. 

Case  II.    Two  sides  and  the  included  angle  are  given. 

Case  III.  Two  sides  and  the  angle  opposite  to  one  of  them 
are  given. 

Case  IV.    All  three  sides  are  given. 

53.   Case  I.     Given  one  side  and  two  angles.     Let  <?,  A^  B 

be  the  given  quantities.     We  may  use  the  formulae 


(1) 


(7=180°-  (^+^), 
c^m  A     7       c  sin  B 


a  = 


h  = 


sin  Q  '  sin  Q 

to  compute  (7,  a,  and  h. 

The  most  reliable  and  convenient  check  is  furnished  by  one 
of  Mollweide's  equations.     (See  Equations  (6),  Art.  51.) 

(2)  ^_^^.sinia-J). 

cos  J  U 

The  law  of  tangents  may  also  serve  as  a  check  instead  of  (2), 
but  it  is  not  quite  as  convenient.  In  calculating  the  check, 
it  is  convenient  to  think  of  A  as  larger  than  B^  so  that  A  —  B 
may  be  positive.  If  A  is  less  than  B^  interchange  A  and  B^ 
and  also  a  and  h  in  equation  (2). 

EXERCISE  XXIV 

Example  1.  Given  c  =  327.85,  A  =  110°  52'.9,  B  =  40'  31'.7.    Find  C, 
a,  and  b.     Check  the  results. 


TWO   SIDES  AND   INCLUDED   ANGLE 


109 


c  =     327.85       (1) 

log  c  =  2.51568 

(6) 

Given    ]  A  =  110°  52'.9    (2) 

log  sin  A  =  9.97049-10 

(7)* 

B=    40^'3r.7    (3) 

colog  sin  C  =  0.32008 

(9)t 

^  4-  ^  =  151°  24'.6    (4) 

log  a  =  2.80625 

(10) 

C=    28°  35' .4    (5) 

a  =  640.10 

(12) 

log  c  =  2.51568 

(6) 

log  sin  B  =  9.81280-10 

(8) 

colog  sin  C  =  0.32008 

(9) 

log  b  =  2.64856 

(11) 

b  =  445.20 

(13) 

Check. 

A-B=70°  21'.2     (14) 

logc  =  2.51568 

(6) 

^(A-B)  =  35°  lO'.Q     (15) 

log  sin  l(A  -B)=  9.76050-10 

(17) 

^  C  =  14°  17'.7     (16) 

colog  cos  I'C  =  0.01366 

(18)  t 

log(a  -b)=  2.28984 

(19) 

a-b  =  194.90      (12) 

-(13) 

a-b  =  194.91   (From 

(19)) 

Remarks.  The  numbers  (1),  (2),  etc.,  indicate  the  order  in  which  the 
separate  results  are  written  down  and  are  meant  to  assist  the  student  in 
understanding  the  arrangement  of  the  computation.  These  numbers 
should  not  appear  in  the  student's  own  work. 


Solve  and  check  the  following  triangles. 


a  =  467.00, 

a  =  24.31, 

a  =  148.30, 
A  =  71°  13'  30", 

a  =  3.4356, 
A  =  47°  13'.2, 


A  =  56°  28'.0, 
A  =  45°  18', 
A  =  37°  24'.0, 
B  =  40°  34'  15" 
A  =  17°  43'.4, 
B  =  65°  24'.5, 


B  =  69°  14'. 
B  =  22°  11'. 
C  =  76°  48'.5. 
c  =  236.54. 
C  =  60°  35'.7. 
a  =  43.176. 


B  =  100°  21'  10",  C  =  58°  17'  20",  a  =  31.656. 
a  =  52.780,  ^  =  37  °41'  15",  B  =  77°  29'  40". 


10.   A  =  57°  23'  12",    C  =  68°  15'  30",  c  =  832.56. 


54.    Case  II.      Given  two   sides  and  the  included  angle. 

Let  a,  5,  and  0  be  the  given  parts.     If  a  >  5,  we  use  the  for- 
mula 


(1) 


(A  +  B)  =  90= 


iO 


i. 


*  Remember  that  sin  110°  52'. 9  =  sin  (90°  +  20°  52'. 9)  =  cos  20°  52'. 9. 
t  Remember  the  rule  for  finding  a  cologarithm.     (Art.  23.) 


(Art.  43.) 


110  SOLUTION   OF   OBLIQUE   TRIANGLES 

to  find  ^  ( J.  4-  ^),  and  the  law  of  tangents, 

(2)  tanlU'-^)=^^tanl(^+^)=^:=4cotl(7, 

to  find  \(^A—B).     We  then  find  A  and  B  from 

(3)  A  =  liA  +  B^  +  ^{A-B\ 
B  =  ^A  +  B)~^iA-B\ 

and  apply  the  law  of  sines  to  find  c,  giving 

(4)  c=^^^. 

sin  J. 

As  checks  we  use  the  relations 

(5)  ^  +  5+ (7=  180°, 

and  one  of  the  Mollweide  equations,  in  the  form 

.g.  a-h  _  sinl(J.- ^) 

^  ^  c     ~^        cosi  O      '     . 

The  law  of  sines  furnishes  another  relation  which  might 
be  used  as  a  check  instead  of  (6).     But  (6)  is  more  reliable. 

If  a  <  6,  we  interchange  the  letters  a  and  6,  and  also  A 
and  B  in  all  of  the  above  formulae,  to  avoid  the  appearance 
of  negative  angles. 

EXERCISE    XXV 

Example  1.  Given  a  =  469.71,  h  =  264.37,  C  =  96°  57'.6.  Find  A,  B, 
and  c  and  check  the  results. 

Solution. 

C  =  96°57'.6  (1)  j^  C  =  48°  28\8  (6) 


Given      a  =  469.71  (2)  logcot^  C  =  9.94711  -  10  (8) 

[  b  =  264.37  (3)  log  (a -Z/)  =  2.31247  (10) 

a +  6^=734.08  (4)  colog(a  +  6)  =  7.13425  -  10  (11) 

a-b  =  205.34  (5)  log  tan  ^(A  -B)=  9.39383  -  10  (12) 

^(A+B)  =  41°  31'.2  (7) 

loga  =  2.67183  (18)  ^  (^  _  b)  =  13°  54'.6  (13) 

log  sin  C  =  9.99679  -  10  (19)  A  =  55°  25'.8  (15) 

cologsin^  =  0.08437  (20)  B  =  27°  36'.6  (16) 

logc  =  2.75299  (21)  C  =  96°  57'.6  (1) 

c  =  566.23 


TWO   SIDES  AND   AN   OPPOSITE   ANGLE 


111 


Checks. 
,log  (a -&)=  2.31247 
logc  1=2.75299 

log^^  =  9.55948 
c 


(10)  log  sin \{A-B)^  9.38091  -  10  (14) 
(21)       logcos^C=:  9.82144 -10   (9) 

10  (22)  logfi^lMA^:?)  3Z  9.55947  -  10  (23) 


cos^  C 

A+B+C  =  180°  O'.O  (17) 

The  checks  consist  in  the  result  (17)  and  the  close  agreement  of  (22) 
and  (23). 

Solve  and  check  the  following  triangles  : 


2. 

b  =  472, 

c  =  324, 

A  =  78°  40'. 

3. 

a  =  748, 

b  =  375, 

C  =  63°  35'.5. 

4. 

a  =  42.38, 

b  =  35.00, 

C  =  43°  14' 40". 

5. 

h  =  0.941, 

c  =  1.256, 

A  =  35°  17'  28". 

6. 

a  =  12.3460, 

b  =  5.7213, 

C  =  65°  30'  10". 

7. 

a  =  25.384, 

c  =  52.925, 

B  =  28°  32'  20". 

8. 

&  =  p.  14367, 

c  =  0.11412, 

A  =  42°  14'.6. 

9. 

a  =  138.65, 

b  =  226.19, 

C  =  59°  12'  54". 

10. 

b  =  1436.7, 

c  =  1141.2, 

A  =  42°  14'  35". 

55.  Case  III.  Given  two  sides  and  the  angle  opposite  to 
one  of  them. 

G-eometric  discussion.  Let  J.,  <x,  h  be  the  given  parts.  We 
construct  the  angle  XA  Y  equal  to  the  given  angle  A,  and 
lay  off  AC,  on  AY^  equal  to  the  given  side  h.  With  /7as 
center,  and  a  radius  equal  to  the  given  side  a,  we  strike  an 
arc.  If  this  arc  intersects  AX  in  B  and  B' ^  one  or  both 
of  the  triangles  ABO  and  AB'  0  may  be  solutions  of  the 
problem. 


The  following  cases  may  present  themselves  if  A  is  an 
acute  angle: 

I.  a<  OP,  i.e.  a  <  5  sin  A     (Fig.  48.) 

The  triangle  is  impossible  in  this  case,  since  the  arc  of 
radius  a,  with  C  as  center,  will  not  intersect  AX. 


112  SOLUTION  OF  OBLIQUE   TRIANGLES 

IL  a=CP  =  5  sin  A     (Fig.  49.) 

The  solution  is  the  right  triangle  APQ. 

III.  h>a>h^inA.     (Fig.  50.) 

There  are  two  solutions  in  this  case.      The  triangles  ABQ 
and  AB'C  both  satisfy  all  of  the  requirements  of  the  prob- 
lem. 

IV.     a'^h.     (Fig.  51.) 

If  a>  b^  there  is  one  solu- 
tion only  ;  namely,  ABO.    The 
Fig.  51  triangle  AB' 0  does  not  con- 

tain the  given  angle  A^  but  its 
supplement.  If  a  =  b,  ABC  is  isosceles,  and  AB' O  reduces 
to  a  triangle  of  zero  area.  Thus  we  may  say  that  there  is 
one  solution  only  ii  a>b. 

The  following  additional  cases  may  present  themselves  if 
A  is  an  obtiise  angle. 


^- 


A  B 

Fia.  52 

V.  a<b.     (Fig.  52.) 

The  triangle  is  impossible  ii  a  <b  and  of  zero  area  if  a  =  5. 

VI.  a>b,     (Fig.  53.) 

There  is  one  solution;  namely,  the  triangle  ABO.  The 
triangle  AB  O  does  not  contain  the  given  angle  ^,  but  its 
supplement. 

Trigonometric  discussion.  How  do  the  equations  of  trigo- 
nometry bring  into  evidence  these  various  cases  ? 

The  law  of  sines  gives 

,1V                                        '     r>     b  sin  A 
(1)  sin  B  = . 


TWO   SIDES   AND   AN   OPPOSITE   ANGLE  113 

If  A^  «,  h  are  giveu,  we  may  compute  the  right  member  of 
this  equation.     Suppose  first  that  A  is  an  acute  angle. 

If  the  right  member  of  (1)  is  greater  than  unity,  the 
triangle  is  impossible,  since  the  sine  of  an  angle  is  never 
greater  than  unity.  This  is  Case  I  of  the  geometric  discus- 
sion. 

If =  1,  5  must  be  a  right  angle.     (Case  II.) 

a 

If  the  given  values  of  a,  5,  and  the  acute  angle  J.,  are  such 

that 

(2)  Jjin^^l^ 

a 

we  can  find  not  merely  one  angle  to  satisfy  equation  (1)  but 
two  such  angles,  which  are  supplementary  to  each  other,  one 
acute  and  one  obtuse.  For  if  K  is  the  acute  angle  which 
satisfies  equation  (1),  the  obtuse  angle  180°  —  K^  which  has 
the  same  sine  as  K  (See  Art.  40),  will  also  satisfy  equa- 
tion (1). 

Thus  we  find  in  the  first  place,  from  (1),  the  two  pos- 
sibilities 

(3)  B  =  K  (acute  angle),   B  =  180°  -  K  (obtuse  angle). 

We  must  remember,  however,  that  A-\-  B  must  be  less  than 
180°.  If,  as  we  have  supposed,  A  is  an  acute  angle,  A-\-  K 
is  certainly  less  than  180°.  But  A  +  180°  —  ^  is  less  than 
180°  if  and  only  '\i  A<  K\  that  is  (see  Fig.  50,  where 
Z.  K=  /.  ABQ^^  if  and  only  if  a<h.  Only  in  this  case 
then  will  there  be  two  solutions.  That  is,  we  have  two 
solutions  if  A  is  acute  and  \ih  >  a  >h  sin  A^  in  accordance 
with  Case  III  of  the  geometric  discussion. 

If  A  is  acute,  but  if  J. -f- 180°  -  Z>  180°;  that  is,  if 
A>K  and  therefore  a>b,  the  obtuse  angle  solution 
180°  —  IC  for  B  becomes  inadmissible  and  the  problem  has 
only  one  solution,  in  agreement  with  Case  IV  of  the  geo- 
metric discussion. 

If  J.  is  an  obtuse  angle,  the  obtuse  angle  solution  180°  —  K 
for  B  is  never  admissible,  since  a  triangle  can   contain  at 


114  SOLUTION   OF   OBLIQUE   TRIANGLES 

most  one  obtuse  angle.  The  acute  angle  solution  B  =  K 
is  admissible  if  and  only  if  A-\-  K  is  less  than  180°.  This 
distinction  gives  rise  to  the  two  remaining  Cases,  V  and  VI, 
of  the  geometric  discussion. 

In  the  trigonometric  solution  of  a  numerical  problem  of 
this  kind,  it  is  essential  to  remember  ithe  following  facts : 

1.  If^on  computing  the  sine  of  an  angle^  we  find  its  value  to 
he  greater  than  unity ^  the  triangle  is  impossible. 

2.  If  the  sine  of  an  angle  is  found  to  be  a  positive  proper 
fraction^  there  are  two  possibilities  for  the  corresponding  angle. 
One  of  these  angles  is  acute  and  the  other^  the  supplementary 
angle,  is  obtuse. 

3.  77ie  sum  of  any  two  angles  of  a  triangle  must  be  less  than 
180°. 

The  sine  of  B  having  been  found  from  the  law  of  sines,  as 
indicated  above,  it  will  become  apparent  from  the  correspond- 
ing values  of  B  whether  the  number  of  solutions  is  0, 1,  or  2. 
If  there  is  one  solution,  we  find  O  from 

A-{-B-\-  (7=180° 

and  c  from  the  law  of  sines.  We  may  check  by  one  of 
Mollweide's  equations  or  by  the  law  of  tangents.  If  both 
values  of  B  are  admissible,  we  use  each  of  them  in  succession, 
so  as  to  find  the  remaining  parts  of  the  two  triangles  which 
are  solutions  of  the  problem. 

EXERCISE  XXVI 

Example  1.  Given  A  =  15^^  32'.7,  a  =  103.21,  b  =  152.37.  Find  the 
remaining  parts  of  the  triangle  or  triangles  determined  by  these  data. 

Solution. 

Formulce :  sin  B  =  ^.mA,  c  =  180°  -(A-{-B),c  =  ^^}J^. 
a  am  A 

Check:  6_a  =  ^iiilii|^. 

COS  l  C 

A  =  15°  32'.7     (1)  fB  =  23°  18'.4        B'  =  156°  41.6 

Given   \  a  =  103.21         (2)         Results    C  =  141°  8'.9        C  =  7°  45.7 
b  =  152.37         (3)  i  c  =  241.58  c'  =  52.01 


TWO  SIDES  AND  AN  OPPOSITE   ANGLE 


115 


Computation. 

log  6  =  2.18290 

(4) 

log  a  =  2.01372 

(5) 

log  sin  A  =:  9.42813 

(7) 

log  sin  C  =  9.79748- 

10 

(15) 

cologa  =  7.98628-10 

(«) 

colog  sin  A  =  0.57187 

(17)* 

logsin5  =  9.59731-10 

(8) 

log  c  =  2.38307 

(18) 

Bz=    23°18'.4 

(9) 

c  =  241.58 

(20) 

B'  =  156°  41'.6      . 

(10) 

]oga  =  2.01372 

(5) 

B  +  A=    38°  5r.l 

(11) 

log  sin  C  =  9.13050- 

10 

(16) 

B'  +  A=  172°  14'.3 

(12) 

colog  sin  ^  =  0.57187 

(17)* 

C  =  Ur    8'.9 

(13) 

log  c'  =  1.71609 

(19) 

C  =     7°45'.7 

(14) 

c'  =  52.01 

(21) 

Check. 

B 

-A  = 

7°45'.7       (22) 

B' 

-A  = 

141°   8'.9        (23) 

KB- 

-A)  = 

3°  52'.9        (24) 

KB>- 

.A)  = 

70°  34'.5        (25) 

log  c  =  2.38307        (18) 

log  sin  K^-^)  =  8.830.56-10  (26) 

colog  cos  ^(7=0.47811        (28) 

(30) 

(32)t 


log  (6-  a)  =  1.69174 
&-a  =  49.17 


log  c'  =  1.71609        (19) 

log  sin  K^'- -4)  =  9.97455-10  (27) 

colog  cos  ^C'  =  0.001 00        (29) 

log(6- a)  =1.69164        (31) 

6-a  =  49.16  (33)t 


5-a  =  49.16 


(34)t 


Example  2.     Given  A  =  15°32'.7,  a  =  10.321,  b  =  152.37.     Find  the 
remaining  parts  of  the  triangle  or  triangles  determined  by  these  data. 

Solution. 


Given 


A  =  15°  32\7  log  b  =  2.18290  - 

a  =  10.321  log  sin  A  =  9.42813-10 

[  b  =  152.37  colog  q  =  8.98628-10 

log  sin  B  =  0.59731 
Since  log  sin  5  has  the  characteristic  zero,  sin  B  is  greater  than  unity. 
Therefore,  the  triangle  is  impossible. 

Example  3.     Given  A  =  15°  32^7,  a  =  167.38,  b  =  152.37.     Find  the 
remaining  parts  of  the  triangle  or  triangles  determined  by  these  data. 

Solution. 


^4  =  15°  32'.7 
Given     a  =  167.38 
b  =  152.37 


\B 
Results     C 


14°   7'.2 
150°  20'.1 
c  =  309.11 


*  Obtained  from  (7) . 


t  Obtained  from  the  logarithm  above. 


^.  Obtained  by  subtraction  from  (2)  and  (3). 


116 


SOLUTION  OF  OBLIQUE   TRIANGLES 


Computation. 

log  h  =  2.18290 

loga  =  2.22371 

log  sin  A  =  9.42813-10 

log 

sin  C  =  9.69454-10 

colog  a  =  7.77629-10 

colog 

sin  A  =  0.57187 

log  sin  B  =  9.38732-10 

log  c  =  2.49012 

B=    14°   7'.2 

c  =  309.11 

B'=165°52'.S 

B-h  A=    29°  39'.9 

B'  +A  =181°25'.5* 

C-150°20M 

Check. 

log  c  =  2.49012 

A  -B=    1°25'.5 

log 

sin  l(A 

-B)  =  8.09516-10 

1{A-B)=    0°42'.8 

colog  cos  1  C  =  0.59180 

1  C  =  75°  lO'.l 

log  (a -/0=  1-17708 

a-b  =  15.01 

< 

2  -  6=15.03 

Find  out  whether  the  triangles  corresponding  to  the  following  data 
are  possible,  how  many  solutions  there  are,  and  what  are  the  values  of 
the  missing  parts. 


4. 

a  =  98, 

b  =  100, 

A  =  120°. 

5. 

a  =  767, 

b  =  242, 

A  =  36°  53'  2''. 

6. 

a  =  3541, 

b  =  4017, 

A  =  61°  27'. 

7. 

a  =  67.53, 

b  =  56.82, 

A  =  77°  14'  19". 

8. 

a  =  9.4672, 

c  =  14.433, 

A  =  IV  14'.3. 

9. 

a  =  413.28, 

b  =  378.19, 

B  =  50°  16'  25". 

10. 

a  =  345.46, 

b  =  531.75, 

A  =  26°  47'  32". 

56.   Case  IV.     Given  the  three  sides  of  the  triangle.     We 

use  the  formulae 


tsinlA= ,   t'diihB  = -,   tan  J  (7= , 

8  —  a  8  —  0  8  —  c 


and  the  check 


A-hB-\- 0=180''. 


A 


♦Therefore  B'  is  inadmissible.    There  is  only  one  solution,  as/might  have 
been  foreseen,  since  a  >  6.  / 


GIVEN  THE   THREE   SIDES 


117 


EXERCISE  XXVII 

Example   1.     Given  a  =  34.278,   b  =  25.691,   c  =  30.175.     Find   the 
angles  A,  B,  C. 

Solution. 

[a  =  34.278 


Given      &  =  25.691 

I  c  =  30.175 

2^=90.144 

.^  =  45.072 

s-a  =  10.794 

s-&  =  19.381 

s-c  =  14.897 

2. s  =  90.144* 

logr=0.91987 

log  (.9- a)  =  1.03318 


logr: 

log  (5 -6): 


\A=  75°13'.0 

Results      B=  46°26'.6 

[  C=   58°  20^2 

Check.     ^+^  +  C=179°59^8 

colog  .9  =  8.34609 -10 
log(s-a)  =  1.03318 
log  (s- 6)  =  1.28737 
]og(g-c)  =  1.17310 
log  r2=  1.83974 1 
:0.91987  logr  =  0.91987t 

1.28737  log  (5 -c)  =  1.17310 


log  tan  M  =  9.88669  - 10  log  tan  ^  5  =  9.63250  - 10  log  tan  ^  C  =  9.74677  - 10 
^  =  37°  36'.5  ^B  =  2^°  13'.3  ^C=29°  lOM 

Remark.  In  this  problem  some  time  may  be  saved  by  writing  log  r  on 
the  lower  margin  of  a  slip  of  paper  and  placing  it  above  log  (s  —  a)  to 
find  log  tan  ^  A,  above  log  (s  —  h)  to  find  log  tan  J  B,  and  above  log  {s  —  c) 
to  find  log  tan  ^  C.  A  similar  device  is  often  useful  in  similar  cases. 
Most  computers  also  save  time  by  omitting  the  — 10  attached  to  logarithms 
with  negative  characteristics.  This  omission  can  never  give  rise  to 
serious  misunderstanding. 

Find  the  angles  of  the"  triangles  whose  sides  have  the  following  values : 


2. 

a  =  79.3, 

b  =  94.2, 

c  =  66.9. 

3. 

a  =  0.785, 

b  =  0.850, 

c  =  0.633. 

4. 

a  =  312, 

b  =  423, 

c  =  342. 

5. 

a  =  25.17, 

b  =  34.06, 

c  =  22.17. 

6. 

a  =  93146, 

b  =  176530, 

c  =  95768. 

7. 

a  =  12.653, 

b  =  17.213, 

c  =  23.106. 

7/  /3y  , 


If  oniy  one'of  the  three  angles  is  to  be  calculated,  it  may  be  more  con- 
venienji  to  make  use  of  the  formulae  for  sin  ^  ^  or  cos  ^  ^,  which  may  be 
found  ^rom  the  indications  given  in  Example  4,  Exercise  XXII. 


*  Obtained  by  adding  s,  s  —  a,  s  —  b,  s  —  c,  as  a  check  on  the  additions  and 
subtractions  required  to  find  these  quantities. 

t  Obtained  by  adding  the  four  logarithms  above  log  r^,  since 
^.._(s-a)(s-b)is-c)_ 


t  Obtained  by  taking  one  half  of  log  r^. 


118  SOLUTION    OF  OBLIQUE   TRIANGLES 

57.  Problems  in  heights  and  distances,  plane  surveying,  and 
plane  sailing.  Some  of  the  following  problems  are  direct  appli- 
cations of  the  methods  which  have  just  been  explained.  In 
others,  it  will  be  necessary  to  consider  several  triangles  in 
succession  or  simultaneously. 

It  will  always  be  advisable  to  draw  a  figure,  approximately 
to  scale,  to  denote  the  known  as  well  as  the  unknown  sides 
and  angles  of  the  figure  by  properly  chosen  letters,  and  to 
write  down  the  formulae  to  be  used  in  their  general  form, 
leaving  the  substitution  of  the  numerical  values  to  the  last. 
Much  blundering  and  much  unnecessary  work  may  be  avoided 
by  adopting  this  plan. 

The  student  should  use  his  judgment  in  regard  to  the 
number  of  decimal  places  used  in  the  numerical  part  of  the 
work.  Use  three-place  tables  whenever  possible.  Many  of 
the  problems  may  be  solved,  wholly  or  in  part,  by  the  slide- 
rule. 

EXERCISE  XXVIII 

1.  In  order  to  find  the  height  of  a  tower,  the  angles  of  elevation  of 
its  top  are  measured  from  two  stations,  A  and  B,  in  the  same  horizontal 
line  with  its  base,  and  on  the  same  side  of  the  tower.  If  the  angles  of 
elevation  of  the  tower  from  A  and  B  are  32°  and  65°  respectively,  and  if 
the  distance  AB  is  500  feet,  find  the  height  of  the  tower. 

2.  Find  the  distances  from  the  two  stations  of  Ex.  1  to  the  foot  of  the 
tower. 

3.  If  the  angles  of  elevation  of  the  tower  from  A  and  B,  in  Ex.  1, 
are  L  and  M  respectively,  and  if  the  distance  AB  is  equal  to  d  feet,  show 
that  the  height  of  the  tower  is 

\  ,  _  d  sin  L  sin  M 

~sin(M-Z)* 

4.  An  obstacle  (a  house)  was  found  to  interfere  with  the  running  of 
a  straight  line  from  A  in  the  direction  AB.     (See  Fig.  54.)     An  angle 

ABE  was  turned  at  B,  equal  to  123°,  and 
the  distance  BE  was  measured  equal  to 
150  feet.  The  angle  BEC  was  made  equal 
to  63°.  How  long  must  the  distance  EC 
be,  and  what  angle  must  be  turned  at  C,  in 
order  that  CD  may  be  the  prolongation  of 
Fig.  54  AB? 


APPLICATIONS  INVOLVING  OBLIQUE   TRIANGLES     119 

5.  How  may  angles  at  B  and  E  be  chosen,  in  Ex.  4,  so  as  to  avoid 
all  computation  ? 

6.  Two  straight  railroad  tracks  intersect  at  an  angle  of  75°.  What 
will  be  the  distance,  at  the  end  of  20  minutes,  between  two  trains  which 
start  from  the  crossing  at  the  same  instant,  their  speeds  being  30  and  40 
miles  per  hour  respectively  ? 

7.  Each  of  two  battleships  passing  each  other  fired  a  salute.  A 
person  on  shore  observed  that  the  interval  between  the  flash  and  the 
report  of  the  gun  was  4  seconds  for  one  ship  and  6  seconds  for  the 
other.  The  angle,  at  his  eye,  subtended  by  the  two  ships,  just  as 
the  salute  was  fired,  was  55°.  The  velocity  of  sound  is  about  1140  feet 
per  second.     Find  the  distance  between  the  ships. 

8.  Two  lighthouses  are  2.789  miles  apart,  and  a  certain  rock  is  known 
to  be  4.325  miles  from  one  of  them.  The  angle  subtended  by  the  two 
lighthouses  at  the  rock  is  16°  13'.  How  far  is  the  rock  from  the  other 
lighthouse?  How  many  solutions  are  there  to  this  problem?  Can  we 
make  a  choice  between  these  solutions  if  we  know  which  of  the  two  light- 
houses is  nearer  to  the  rock  ? 

9.  Find  the  radius  of  the  largest  cylindrical  gas  tank  which  can  be 
constructed  on  a  triangular  lot  whose  sides  measure  73,  82,  91  feet  respec- 
tively, and  locate  the  center  of  its  circular  base. 

10.  An  observer  measures  the  angle  of  elevation  of  a  cloud  due  south 

of  him  at  the  moment  when  the  sun  also  is  due  south  (at  apparent  noon).       ^ 
The  angle  of  elevation  of  the  sun  was  65°,  that  of  the  cloud  75".     If  the      V 
shadow  of  the  cloud  falls  550  feet  north  of  the  observer,  how  high  is  the 
cloud  ? 

11.  At  9  P.M.  two  lights,  known  to  be  8  miles  apart,  are  observed  to 
be  due  east  from  a  certain  vessel.  At  10  p.m.  one  of  these  lights  bears 
N.E.  and  the  other  N.N.E.  If  the  course  of  the  ship  was  due  south,  what 
was  its  rate  ? 

12.  A  tower  is  situated  on  top  of  a  conical  hill  whose  sides  make  an 
angle  of  15°  with  the  horizontal  plane.  At  a  distance  of  120  feet  from 
the  foot  of  the  tower  (the  distance  being  measured  along  the  slope)  the 
tower  subtends  an  angle  of  20°.     Find  the  height  of  the  tower. 

13.  If,  in  Ex.  12,  the  side  of  the  hill  makes  an  angle  /  with  the  hori- 
zontal plane,  and  if  the  angle  subtended  by  the  tower,  at  a  distance  of  d 
feet  from  its  foot,  is  A,  show  that  the  height  of  the  tower  is 

c?  sin  ^ 


A  = 


cos  {A  +  /) 


120 


SOLUTION   OF   OBLIQUE   TRIANGLES 


14.  A  tower  54  feet  high,  situated  on  top  of  a  conical  hill,  subtends  an 
angle  of  15°  30'  at  a  point  120  feet  from  the  foot  of  the  tower  (the 
distance  being  measured  along  the  slope).  What  angle  does  the  side  of 
the  hill  make  with  the  horizontal  plane  ? 

15.  To  find  the  slope  of  a  railroad  embankment,  one  end  of  a  pole  12 
feet  long  was  placed  on  the  level  ground  6  feet  from  the  foot  of  the  em- 
bankment, and  the  other  end  was  found  to  fall  at  a  point  7.5  feet  up  its 
face.     What  angle  does  the  embankment  make  with  a  horizontal  plane  ? 

Remark.     A  transit  cannot  conveniently  be  used  to  measure  an  angle 
formed  by  two  walls,  the  angle  formed  by  an  embankment  or  buttress 
with  a  horizontal   plane,  etc.     In  such  cases,  as  in  this  example,  it  is 
more  convenient  to  measure  distances  and  deter- 
mine the  angles  by  calculation. 

16.  Two  capes,  A  and  B  (Fig.  55),  were  ob- 
served from  a  ship  at  sea;  one  of  them  bore 
N.N.E.  and  the  other  N.W.  It  was  found  from 
the  chart  that  the  second  cape  bore  W.  by  N. 
from  the  first  and  was  25.3  miles  distant  from  it. 
What  was  the  distance  of  the  ship  from  each  of 
the  two  capes  ? 

17.  A  battleship  leaves  port  ^,  on  a  due  easterly  course,  at  the  rate  of 
16  miles  per  hour.  A  dispatch  boat  starts  from  B  at  the  same  moment. 
The  port  B  bears  S.S.W.  of  port  A  and  is  25 

miles  distant  from  it.  If  the  dispatch  boat  has  a 
rate  of  22  miles  per  hour,  what  should  be  the 
direction  of  its  course  so  that  it  may  meet  the 
battleship,  if  neither  ship  alters  its  rate  or  course  ? 
At  what  time  will  they  meet  ? 

Hint.     In  Fig.  56  we  have  AC  =  Ut,  BC  = 
22  t,  if  t  denotes  the  time  (in  hours)  which  passes 

between  the  time  of  sailing  and  the  moment  of  meeting,  and  if  C  repre- 
sents the  place  of  meeting. 

18.  The  angle  of  elevation  of  the  top  of  a  tower,  at  a  point  in  the 
same  horizontal  plane  with  its  base,  is  equal  to  ^.  At  a  point  h  feet 
directly  above  the  first  the  angle  of  depression  of  the  foot  of  the  tower 

was  found  to  be  equal  to  B.  Prove  that 
the  height  of  the  tower  is  equal  to  h  tan  A 
cot  B. 

19.  A  valley  has  the  cross  section  shown  in 
Fig.  57,  the  angles  L  and  M  and  the  distance 
AB  =  1  having  been  obtained  by  a  survey.  It 
is  planned  to  connect  the  points  A  and  JB  by  a 


Fig.  56 


APPLICATIONS  INVOLVING  OBLIQUE   TRIANGLES      121 

bridge,  supported  by  a  pier  at  C.     How  high  must  this  pier  be  made? 

20.  Show  that  the  area  of  a  quadrilateral  is  \dd'  sin  A^ii  d  and  d' 
are  the  lengths  of  its  diagonals  and  if  A  is  one  of  the  angles  which  the 
diagonals  make  with  each  other. 

21.  Two  sides  of  a  parallelogram  are  3.41  and  2.60  feet  long  and  the 
shorter  diagonal  is  1.58  feet  long.     Find  the  length  of  the  other  diagonal. 

22.  The  sides  of  a  field  ABCD  2^re.  AB  =  b1  feet,  5C  =  43  feet,  CD =45 
feet,  DA  =  47  feet ;  and  the  distance  from  ^  to  C  is  50  feet.  Find  the 
area  of  the  field. 

23.  Two  streets  intersect  at  an  angle  of  75°.  The  corner  lot  has 
frontages  of  150  feet  and  115  feet  on  the  two  streets,  and  the  remaining 
two  boundary  lines  of  the  lot  are  perpendicular  to  the  two  streets.  What 
is  their  length,  and  what  is  the  area  of  the  lot  ? 

24.  In  order  to  measure  the  distance  between  two  pumping  stations, 
A  and  B,  in  Lake  Michigan,  a  base  line  CD  =  11.1  chains  was  measured 
along  the  shore.  (See  Fig.  58.)  The  follow-  ^  ^  B 
ing  angles  were  measured  : 

^Ci)=  Ci  =  132°29',  _ 

ACB  =  82°  20',  ^^P ^:^^^^'^H 

CDA  =Di  =  45°  59', 

CZ>5  =  A  =  124°48.  .^'9   ■•'       -1^.-7 

Fig.  58     ' 
Compute  the  distance  AB. 

(Fig.  58  is  not  drawn  to  scale.) 

25.   Devise  a  plan  for  finding  the  distance  between  any  two  inacces- 
sible points,  A  and  B,  in  the  same  horizontal  plane  if  two  points,  C  and 
D,  can  be  found  in  the  same  plane,   from 
both  of  which  A  and  B  are  visible. 

26.  Devise  a  plan  for  finding  the  dis- 
tance between  two  inaccessible  points,  A 
and  B,  if  both  are  visible  from  only  one  ac- 
cessible point  C. 

Hint.  In  Fig.  59,  select  a  point  D  from 
which  A  and  C  are  visible,  and  a  point  E  from  which  B  and  C  are 
visible.     Measure  CD,  CE,  and  the  angles  D,  Ci,  C2,  C3,  E. 

27.  To  compute  the  distance  between  two  accessible  points,  A  and  B, 
if  no  point  can  be  found  from  which  both  A  and  B  can  be  seen.  (For 
instance,  if  A  and  B  are  points  on  opposite  sides  of  an  inaccessible 
mountain.)  Take  a  point  C  from  which  A  may  be  seen  and  a  point  D 
from  which  B  is  visible.  If  C  is  visible  from  D,  measure  the  angles 
ACD  and  CDB  and  the  distances  AC,  CD,  DB.  Show  how  to  calcu- 
late tlie  distance  AB  from  these  data. 


122 


SOLUTION   OF   OBLIQUE   TRIANGLES 


Fig.  60 


28.   If  the  points  A  and  B  of  Ex.  27  are  inaccessible,  the  distances 
A  C  and  BD  cannot  be  found  by  direct  measurement.     In  such  a  case 

(see  Fig.  60),  select  points  C,  D,  E,  F  so 
that  A,  C,  E  shall  be  visible  from  D,  and 
D,  F,  B  from  E.  Measure  the  angles 
C,  Di,  Z>25  El,  E2,  F,  and  the  distances 
CD,  DE,  and  EF.  Show  how  to  find  the 
distance  AB  from  these  measurements. 

29.  A  tower  is  situated  on  top  of  a 
conical  hill  as  in  Ex.  12.  Two  points 
A  and  A'  are  chosen  on  the  side  of  the 
hill  at  distances  c?  =  43  feet  and  d'  =  98 
feet  respectively  from  the  top,  the  points'  being  the  lower  one  and 
the  distances  d  and  d'  being  measured  along 
the  slope.  The  angles  subtended  by  the 
tower  at  A  and  A'  were  A  =  42°,  A'  =  23°  re- 
spectively. Find  the  height  of  the  tower  and 
the  angle  of  inclination  of  the  side  of  the  hill. 

30.  Two  observers,  A  and  B,  are  3  miles 
apart,  A  being  due  west  of  B  and  in  the 
same  horizontal  plane.  Both  observers  meas- 
ure, at  the  same  instant,  the  bearing  and 
angle  of  elevation  of  a  balloon.     A  finds  that 

the  balloon  bears  N.  47°  E.  and  that  its  angle  of  elevation  is  23°.  B  finds 
that  the  balloon  bears  N.  SS""  W.  and  that  its  angle  of  elevation  is  19°.5. 
Find  the  height  of  the  balloon  above  the  horizontal  plane  of  A  and  B. 

31.  To  find  the  horizontal  distance  AD  and  the  vertical  distance  DC 
from  A  to  an  inaccessible  point  C  (Fig.  62)  when  it  is  not  convenient 

to  measure  a  base  line  in  the 
same  vertical  plane  with  C. 

Measure  a  horizontal  base 
line  AB,  k  feet  long,  in  any 
direction  through  A.  Let  D  be 
the  foot  of  the  perpendicular 
from  C  to  the  horizontal  plane, 
MN,  of  AB.  Measure  the  hori- 
zontal angles  BAD  =  A  and 
ABD  =  B,  and  the  vertical  an- 
gles DAC  =  A'  and DBC  =  B'. 

k  sin  A 


Fig.  61 


Fig.  62 


Show  that 


AD  = 


h  = 


ksin  B 
sin(^  +  5)' 
A:  sin  B  tan  A' 


^^      sin  (^+5) 
k  sin  A  tan  B' 


sin  {A  +  B)         sin  {A  +  B) 


APPLICATIONS  INVOLVING  OBLIQUE   TRIANGLES      123 


The  two  formulse  for  h  should  give  the  same  result  in  any  numerical 
problem  of  this  kind.  Lack  of  agreement  indicates  inaccuracy  in  one 
of  the  observed  angles  or  in  the  computation. 

32.  Two  points,  A  and  B,  in  the  same  horizontal  plane  and  separated 
by  a  ridge,  are  to  be  connected  by  a  straight  level  tunnel.  In  order  to 
find  the  distance  between  thern,  the  surveyors  measured  the  inclined 
angle,  BCA,  subtended  by  A*B  from  the  top  C  of  a  neighboring  hill, 
whose  height,  CD  =  h,  above  the  plane  of  A  and  B  is  known.  They 
also  measured  the  angles  of  depression  of  A  and  B  from  C.  Devise  a 
method  for  computing  the  length  of  the  tunnel  AB. 

33.  Prove,  by  means  of  the  trigonometric  formulse,  that  the  angles  of 
a  triangle  can  be  found  when  the  ratios  of  the  three  sides  of  the  triangle 
are  given,  even  if  the  absolute  values  of  the  three  sides  are  not  known. 
Prove  the  same  fact  by  geometry. 

34.  Let  ^,  i?,  C  be  three  points  of  a  horizontal  line,  whose  mutual 
distances,  AB  =  b,  AC  =  c,  BC  =  b  -\-  Cy  are  known.  (See  Fig.  63.)  To 
find  the  horizontal  distances,  p,  q,  r,  of  an  inaccessible  point,  E,  from 
these  three  points,  and  the  height  h,  it  suffices  to  measure  the  three  angles 
of  elevation.  A,  B,  C,  of  E  from 
A,  B,  C  respectively. 

For  the  figure  gives 
(1)    p  =  hcotA,        q  =  hcotB, 

r  =  h  cot  C, 
and  also 

q^=:b'^  +  p^-2bp  cos  BAD, 
r^  =  c'^  +  p'^  +  2cp  cos  BA  D ; 
whence 


Substitute  the  values  of  p,  q,  r  from  (1)  and  solve  the  resulting  equa- 
tion for  h^.     We  find 

(2)  A2  =  ^^(^  +  ^) 

^^  (cot^  C  -  cot^  A)b  +  (cof^B-cot^  A)c 

After  computing  h  from  (2),  p,  q,  and  r  may  be  found  from  (1). 

35.  Given  the  mutual  distances  of  three  points  A,  B,  C,to  find  the 
distances  AD,  BD,  CD  from  a  fourth  point  D  in  the  plane  of  ABC  to 
each  of  the  given  three  points,  when  the  angles  are  given  which  the  lines 
AB,  BC,  CA  subtend  at  D. 

This  problem,  usually  called  Pothenofs  problem,  may  be  solved  as  fol- 
lows : 


'Ua^^S)  T; 


124 


SOLUTION   OF   OBLIQUE   TRIANGLES 


In  Fig.  64,  let 

BC  =  a,     CA  =h,     AB  =  c 

be  the  given  mutual  distances  of  A,  B,  and  C.     Of  course  the  angles  of 
the  triangle  ABC  may  then  also  be  regarded  as  known. 

Let    Z  CDB  =  L,     Z.  CD  A  =  M,     A  BDA  =  N 

be  the  angles  subtended  by  the  sides  of  the  triangle 
from  D.  These  angles'  we  regard  as  known  by  meas- 
urement and,  of  course,  N  =  M  -\-  L. 

The  problem  of  finding  the  distances  ADy  BD,  CD 
may  evidently  be  regarded  as  solved,  if  we  can  find 
the  two  angles 

Z  CAD  =  P  and  Z  CBD  =  Q. 

We  shall  show  how  to  compute  these  angles. 

Applying  the  law  of  sines  to  the  triangles  A  CD 
and  BCD,  we  find 

b  sin  P      a  sin  Q 


(1) 


whence 


CD  = 


sinM 
sinP 


sinL  ' 
asmM 


Ct 


A- 


sin  Q      &  sin  Z 
Hence,  by  the  theory  of  proportion, 

sin  P  —  sin  Q  _  a  sinM 


b  sin  L 


sin  P  +  sin  Q      a  sin  M  -{-  bainL 

which  gives  (Equations  (8),  Art.  49), 

1  - 
(2)  '     tanKP-Q)-, 


tanl(P+Q)      1^ 


b  sinL 
a  sin  M 
b  sin  L 


a  sin  M 
On  the  other  hand  we  have 

P+Q+C  +  iV^=  360°, 
and  therefore 
(3)  i(P  +  Q)  =  180°  -  KC  +  N). 

Since  the  angles  C  and  N  are  known,  (3)  gives  the  value  of  ^(P  +  Q). 
If  this  value  be  substituted  in  (2),  we  may  obtain  from  (2)  the  value  of 
j(P  _  Q).  From  l(P  +  Q)  and  |(P  -  Q),  we  find  P  and  Q  themselves 
by  addition  and  subtraction.  The  distance  CD  may  then  be  computed 
by  (1)  in  two  ways,  providing  a  check  for  the  correctness  of  the  work. 

Remark  1.  The  problem  obviously  becomes  indeterminate,  if  the 
point  D  should  happen  to  be  on  the  circumference  of  the  circle  deter- 
mined by  the  three  points  A,  B,  C.  For,  as  the  point  D  moves  along 
this  circumference,  the  angles  L  and  M  do  not  change,  so  that  the  posi- 
tion of  the  point  D  on  the  circumference  is  not  determined  by  the  value 


APPLICATIONS   INVOLVING   OBLIQUE   TRIANGLES      125 

of  .these  angles.  It  is  evident,  then,  that  if  the  point  7)  is  close  to  the 
circumference  of  the  circle  circumscribed  about  the  triangle  ABC,  its 
position  cannot  be  determined  by  this  method  with  any  considerable 
degree  of  accuracy- 

Remark  2.  The  name  Pothenot's  problem  cannot  be  justified  his- 
torically. A  complete  solution  of  this  problem  was  given  by  Snellius 
in  his  Doctrince  triangulorum  canonicce  libri  quatuor,  which  appeared  in 
1627,  almost  seventy  years  before  Pothenot  presented  his  solution  of  the 
same  problem  to  the  Paris  Academy  of  Science. 

36.  Show  that  the  problem  of  Pothenot  may  also  be  solved  by  draw- 
ing a  circle  through  the  three  points  A,  B,  D,  and  by  making  use  of  the 
triangle  ABE,  where  E  is  the  second  point  of  intersection  of  CD  with 
this  circle. 

37.  In  surveying  a  harbor,  a  submerged  rock  was  located,  for  chart- 
ing purposes,  by  sighting  three  known  objects  A,  B,  C  on  land  from  a 
boat  immediately  above  the  rock.  The  known  distances  were  BC  =  a 
=  312  feet,  CA  =b  =  520  feet,  and  the  angle  C  was  known  to  be  65^  27'. 
The  angles  obtained  by  observation  were  L  =  CDB  =  23°  25'  and 
M  =  CD  A  =  32°  52'.  Find  and  check  the  distance  from  the  rock  to  C 
and  the  angle  which  this  line  makes  with  the  side  AC  ot  the  known 
triangle. 

In  Exs.  38  to  45,  we  use  the  notations  of  Chapter  VII;  A,  B,  C  for 
the  angles,  a,  b,  c  for  the  sides,  2  s  for  the  perimeter,  r  and  R  for  the 
radii  of  the  inscribed  and  circumscribed  circle  respectively,  and  S  for 
the  area  of  the  triangle.  Show  how  to  find  all  of  the  sides  and  angles 
of  the  triangles  determined  by  the  following  data. 

38.  a  -^  b,  c,  A  —  B  are  given.  42.   r,  A,  B  are  given. 

39.  a  -  b,  c,  A  —  B  are  given.  43.   S,  A,  B  are  given. 

40.  R,  a,b  are  given.  44.    aS^,  a,  b  are  giVen. 

41.  R,  A,  B  are  given.  45.   s,  R,  a  are  given. 

46.  A  furnace  maker  receives  an  order  for  a  number  of  furnaces,  some 
40  inches  and  some  42  inches  in  diameter.  These  furnaces  are  to  be  fitted 
on  the  outside  with  an  iron  casting  whose  inside  length,  measured  along 
the  arc,  is  26  inches.  In  order  to  avoid  the  necessity  of  making  two  dif- 
ferent castings,  the  manufacturer  considers  the  possibility  of  making  a 
single  casting,  to  fit  exactly  a  furnace  41  inches  in  diameter  but,  of  course, 
not  fitting  exactly  either  of  the  sizes  ordered.  His  experience  tells  him 
that  such  a  casting  will  serve  the  purpose,  if  no  point  of  its  inner  surface 
is  more  than  a  quarter  of  an  inch  from  the  outer  surface  of  the  furnace 
after  being  placed  in  position.  Will  it  be  necessary  to  make  separate 
castings  for  the  two  different  sizes  ? 


126 


SOLUTION  OF  OBLIQUE   TRIANGLES 


58.  Displacements,  velocities,  and  forces.  If  a  body  is 
transported  from  one  place  in  the  plane  to  another,  and  we 
wish  to  describe  its  change  of  position  or  displacement^  it  is 

clearly  not  sufficient  to  state  how 
far  the  body  has  been  moved.  We 
must  also  include  in  our  descrip- 
tion a  statement  concerning  the  di- 
rection of  the  displacement. 


N 


w- 


s 

Fig.  65 


Thus,  the  displacement  from  0  to  P 
(Fig.  65)  may  be  described  completely  by 
stating  its  magnitude  (the  length  of  the 
line-segment  OP),  and  its  direction  (either 
the  bearing  NOP  of  the  line  OP  or  the 
angle  EOP  or  some  other  angle  which 
fixes  the  direction  of  OP). 


If  the  displacement  is  in  a  horizontal  plane  and  the  direc- 
tions from  0  to  iV,  S^  E^  TTin  Fig.  Qb  represent  north,  south, 
east,  and  west  respectively,  the  projections  OP^  and  OP"  of 
OP  on  OE  and  ON  are  called  the  easterly  and  northerly 
components  of  the  displacement.  If  the  displacement  is  not 
horizontal,  we  define,  in  a  similar  manner,  its  horizontal  and 
vertical  components. 

Let  us  suppose  that  a  point  M  is  displaced  from  0  to  P. 
(See  Fig.  66.)  The  displacement  may  be  represented  in 
magnitude  and  direction  by  the  directed  line-segment  OP, 
(The  arrowhead  indicates  that  the  displace-  ^ 

ment  is  from  0  toward  P,  and  not  from  P 
toward  0.)  Let  OQ  represent  a  second 
displacement.  If  the  point  M^  originally 
at  0,  be  made  to  undergo  both  of  these  dis- 
placements in  succession  (in  either  order), 
it  will  ultimately  arrive  at  i2,  where  R  is  the  fourth  vertex 
of  the  parallelogram  determined  by  OP  and  OQ.  For  this 
reason,  the  displacement  OR  is  said  to  be  the  resultant  of 
the  displacements  OP  and  OQ. 

We  may  think  of  the  two  displacements  OP  and  0§  as 
taking  place  simultaneously.     An   instance  of  this  sort  is 


Fig.  66 


DISPLACEMENTS,   VELOCITIES,   AND   FORCES        127 

furnished  by  a  passenger  shifting  his  position  on  board  of  a 
moving  ship.  His  total  displacement,  in  space,  is  the  result- 
ant of  that  which  is  due  to  the  motion  of  the  ship  and  of 
that  caused  by  his  own  muscular  efforts. 

The  velocity  of  a  train,  of  a  bullet,  or  of  any  uniformly 
moving  object,  is  measured  by  the  distance  which  it  describes 
in  a  unit  of  time,  that  is,  by  a  displacement.  Therefore  a 
velocity,  like  a  displacement,  has  direction  as  well  as  magni- 
tude. The  case  of  a  passenger's  stroll  on  board  a  moving 
ship  suffices  to  illustrate  the  phrase :  resultant  of  two  velocities. 
Since  the  velocity  of  a  uniformly  rdoving  body  is  its  dis- 
placement in  a  unit  of  time,  the  resultant  of  two  velocities  is 
found  by  the  same  method  as  a  resultant  of  two  displace- 
ments, i.e.  by  the  parallelogram  construction  illustrated  in 
Fig.  66. 

It  is  one  of  the  fundamental  facts  of  mechanics  (proved 
by  countless  experiments)  that  two  forces  acting  upon  the 
same  material  point  combine  into  a  single  resultant  force  accord- 
ing to  this  same  parallelogram  law.  This  fact  is  generally 
known  as  the  law  of  the  parallelogram  of  forces. 

Owing  to  the  fact  that  displacements,  velocities,  and  forces 
are  directed  quantities  which  combine  according  to  the 
parallelogram  law,  these  three  classes  of  things  have  many 
properties  in  common.  There  are  many  other  instances  of 
quantities  of  this  same  character  and,  on  account  of  their 
importance,  they  have  received  a  special  name. 

Directed  quantities^  which  combine  in  accordance  with  the 
parallelogram  law,  are  called  vectors.* 

By  a  proper  choice  of  the  units,  every  vector  may  be 
represented  by  a  directed  line-segment  or,  what  amounts  to 
the  same  thing,  by  a  displacement. 

Thus,  the  line-segments  of  Fig.  66  may  be  interpreted  as  forces,  if  the 
directions  of  these  line-segments  coincide  with  the  directions  of  the  forces, 
and  if  each  segment  is  made  to  contain  as  many  length  units  as  there  are 
force  units  in  the  corresponding  force. 

*  From  the  Latin  vector,  meaning  one  who  carries  or  conveys. 


128  SOLUTION  OF   OBLIQUE   TRIANGLES 

EXERCISE  XXIX 

1.  A  steamer  is  moving  N.N.E.  with  a  velocity  of  16  miles  pei 
hour.     Find  the  northerly  and  easterly  components  of  its  velocity. 

2.  A  horizontal  force  of  10  lb.  and  a  vertical  force  of  24  lb.  are  acting 
simultaneously  on  a  point.  Find  the  magnitude  and  direction  of  the 
resultant. 

-  3.  A  schooner  is  sailing  due  west  at  the  rate  of  6  miles  per  hour.  A 
sailor  is  crossing  the  deck,  from  south  to  north,  at  the  rate  of  3  miles  per 
hour.    What  is  the  magnitude  and  direction  of  his  velocity  in  space  ? 

4.  A  force  of  250  lb.  is  acting  on  a  body  in  a  direction  which  makes 
an  angle  of  17°  with  the  horizontal  plane.  How  much  of  this  force  tends 
to  lift  the  body,  and  what  part  of  it  tends  to  move  the  body  in  a  hor- 
izontal plane  ? 

5.  Two  forces,  of  magnitudes  350  lb.  and  510  lb.,  respectively,  act 
upon  the  same  point,  in  directions  which  make  an  angle  of  35°  with  each 
other.  Find  the  magnitude  of  the  resultant,  and  the  angles  which  it 
makes  with  each  of  the  component  forces. 

6.  A  force  of  216  lb.  is  resolved  into  two  components,  which  make 
angles  of  27°  and  32°  respectively,  with  the  direction  of  the  original 
force.     Find  the  magnitude  of  each  component. 

7.  A  man  wishes  to  reach  a  point  on  the  opposite  side  of  the  river, 
250  yards  upstream.  The  velocity  of  the  current  is  2.5  miles  per  hour 
and  the  width  of  the  river  is  300  yards.  If  the  man's  rate  of  rowing 
(in  still  water)  is  4  miles  per  hour,  in  what  direction  must  he  point  the 
head  of  the  boat  in  order  that  his  course  may  be  a  straight  line  ? 

59.  Reflection  and  refraction  of  light.  The  path  of  light, 
in  a  homogeneous  medium  like  air,  is  rectilinear.  But  if  a 
ray  of  light  meets  the  polished  surface  of  a  sheet  of  metal  or 
glass,  its  direction  is  changed  in  accordance  with  the  law 
that  the  angle  of  incidence  is  equal  to  the  angle 
of  reflection. 

This  law  is  illustrated  in  Fig.  67,  where 
the  ray  A  0  strikes  the  reflecting  surface  of 
the  mirror  at  0  and  is  Reflected  in  the  direc- 
tion OB.  The  line  OiV,  perpendicular  to 
the  reflecting  surface  at  0,  is  called  its  nor- 
mal. The  angle  NOA,  or  i,  is  called  the 
angle  of  incidence^  and  the  angle  NOB,  or  r,  is  the  angle  of 
reflection.     According  to  the  law  of  reflection  of  light  (veri- 


REFLECTION  AND   REFRACTION   OF  LIGHT 


129 


fied  by   thousands  of  experiments)    these   two   angles   are 
equal. 

When  a  ray  of  light  AO,  after  passing  through  air,  meets 
the  bounding  surface  of  a  second  transparent  medium  like 
glass  (see  Fig.  68),  only  a  part  of  the 
light  is  reflected.  Another  portion 
of  the  light  enters  the  second  medium 
and  continues  on  its  way  in  a  path 
OB^  which  makes  a  certain  angle  with 
the  direction  AO  oi  the  original  ray. 

If  NN'  is  the  perpendicular  or  nor- 
mal to  the  bounding  surface  at  0,  the 
angle  NOA^  or  ^,  is  called  the  angle 
of  incidence   and  N'OB^    or   r,   is  the   angle   of  refraction. 

As  a  result  of  numerous  experiments,  it  has  been  found 
that  the  quotient 


Fig.  68 


sin 


sm  r 


will  have  the  same  value,  for  a  given  kind  of  glass,  for  all 
different  values  of  ^.  In  other  words,  as  the  angle  of  inci- 
dence changes  the  angle  of  refraction  also  changes,  but  in 
such  a  way  as  to  leave  the  quotient 


(1) 


sin  ^ 

sinr 


=  n 


unchanged.     This  quotient  n  is  called  the  index  of  refraction 

of  the  glass  with  respect  to  air.  Its  value  is  different  for 
different  kinds  of  glass.  For  ordinary  crown  glass  n  is 
about  equal  to  1.5. 

If  the  ray  of  light  again  emerges  into  air,  fJs  indicated  in 
Fig.  68,  after  having  passed  through  a  sheet  of  glass  with 
exactly  parallel  sides,  careful  measurements  show  that  the 
ray  5(7  is  parallel  to  the  original  ray  AO.  In  other  words, 
the  direction  of  a  ray  of  light  is  not  changed  by  passing 
through  a  sheet  of  plate  glass  whose  two  faces  are  exactly 
parallel  to  each  other.     Therefore  :  the  index  of  refraction  of 


130 


SOLUTION  OF   OBLIQUE   TRIANGLES 


air  with  respect  to  glass  is  the  reciprocal  of  the  index  of  refrac- 
tion of  glass  with  respect  to  air. 

The  law  of  the  refraction  of  light  was  first  discovered  by 
Snellius  in  1618.  The  simple  formula  (1)  was  given  by 
Descartes  in  1637. 

EXERCISE  XXX 

1.  A  light  is  placed  on  the  line  perpendicular  to  a  plane  circular 
mirror  at  its  middle  point.  The  distance  from  the  light  to  the  mirror  is 
15.76  inches,  and  the  mirror  is  8.32  inches  in  diameter.  Consider  two 
rays  which  strike  the  mirror  in  the  extremities  of  one  of  its  diameters. 
What  angle  will  they  make  with  each  other  after  reflection  ? 

2.  In  an  experiment,  a  source  of  light  is  to  be  placed  at  ^,  a  mirror 
at  B,  and  a  photographic  plate  at  C.  The  distances  are:  AB  =  5.367 
meters,  BC  =  6.329  meters,  CA  =  7.361  meters.  What  angle  must  the 
mirror  at  B  make  with  the  line  AB  so  that  the  light,  reflected  at  B,  may 
pass  through  C  ? 

3.  Two  billiard  balls,  A  and  B,  have  been  placed  at  distances  a  and  b 
inches  respectively  from  the  same  cushion.  The  line  joining  them 
makes  an  angle  L  with  the  cushion.  Let  K  be  the  angle  at  which  the 
first  ball  must  strike  the  cushion,  so  as  to  hit  the  second  after  rebound- 
ing.    Show  that 

taniir  =  ^-±^tani. 
b-a 

Remark.  Billiard  balls  not  endowed  with  a  lateral  rotation  (without 
"  English  ")  rebound  in  accordance  with  the  law  of  reflected  light. 

4.  Find  the  index  of  refraction  from  the  following  observations. 

(WuUner) 


Angles  of  incidence,  i     .     .     . 

40'^ 

60° 

80° 

Angles  of  reffaction,  r   .     .     . 

24°  24' 

33°  38' 

38°  57' 

The  three  values  obtained  for  n  will  disagree  slightly  owing  to  inac- 
curacies in  the  measurements. 

5.  A  ray  of  light  strikes  a  plate  of  crown  glass  at  an  angle  of  inci- 
dence of  37°.  Find  the  angle  between  the  reflected  and  the  refracted 
ray,  if  the  index  of  refraction  is  1.559. 


REFLECTION  AND  REFRACTION  OF  LIGHT 


131 


Fig.  69 


6.  A  ray  of  light  ABCD  passes  through  a 
glass  prism  whose  cross  section  (see  Fig.  69) 
is  an  equilateral  triangle.  If  the  index  of 
refraction  is  1.559,  what  must  be  the  angle 
of  incidence  in  order  that  the  path  of  the 
light  in  the  prism  may  be  parallel  to  one 
of  its  faces?  What  angle  will  the  ray  CD  make  with  its  original  direc- 
tion AB,  after  emerging  from  the  prism? 

,^_^  7.  A  ray  of  light,  ABC,  etc.,  enters  a  glass  prism, 
whose  cross  section  is  an  isosceles  right  triangle  and 
whose  index  of  refraction  is  1.5,  at  the  point  B 
(Fig.  70),  at  right  angles  to  the  face  of  the  prism. 
At  C  no  part  of  the  ray  can  be  refracted  (Why  ?), 
and  all  of  it  is  reflected  in  the  direction  CE.  Such 
Fig.  70  a  prism  is  called  a  total  reflecting  prism. 


THE   GREEK    ALPHABET 


a, 

A 

Alpha 

V, 

N 

N.u 

A 

B 

Beta 

I 

5 

Xi 

7. 

r 

Gamma 

0, 

O 

0  micron 

5, 

A 

Delta 

TT, 

n 

Pi 

€, 

E 

Epsilon 

P^ 

p 

Rho 

r, 

z 

Zeta 

0-,  9 

,2 

Sigma 

'?» 

H 

Eta 

T, 

T 

Tau 

e,  n 

,@ 

Theta 

y, 

T 

Upsilon 

t, 

I 

Iota 

<i>^ 

c|> 

Phi 

/c, 

K 

Kappa 

x^ 

X 

Chi 

X, 

A 

T  /ambda 

^, 

^ 

Psi 

/x, 

M 

Mu 

G), 

a 

Omega 

182 


PART   TWO 

PROPERTIES   OF   THE   TRIGONOMETRIC 
FUNCTIONS 

CHAPTER  IX 

THE  GENERAL   ANGLE  AND   ITS  TRIGONOMETRIC 
FUNCTIONS 

60.  The  notion  of  the  general  angle.  In  elementary  geome- 
try we  usually  think  of  an  angle  as  ready-made.  We  there 
think  of  two  lines  as  given  and  understand  by  the  angle  be- 
tween them  a  measure  of  their  difference  of  direction.  But 
many  reasons  urge  us  to  oppose  to  this  static  idea  what  might 
be  called  the  dynamic  concept  of  angle,  which  presents  an 
angle,  not  as  the  ready-made  difference  of  direction  between 
two  fixed  lines,  but  as  something  which  is  generated  by  the 
rotation  of  a  straight  line  around  a  fixed  point  as  pivot. 

Thus,  for  instance,  we  shall  say  that  the  minute  hand  of  a  clock 
describes,  or  generates,  an  angle  of  90°  in  fifteen  minutes,  an  angle  of 
360°  in  one  hour,  an  angle  of  1890°  in  five  hours  and  a  quarter.  Although 
the  minute  hand  points  to  the  same  place  on  the  face  of  the  clock  after 
any  number  of  complete  revolutions,  we  are  not  likely  to  make  the 
error  of  ignoring  these  complete  revolutions.*  If  we  did,  we  should  be 
ignoring  the  distinction  "between  1  o'clock,  2  o'clock,  3  o'clock,  etc.  If 
we  were  to  say  that  an  angle  of  360°  is  the  same  as  one  of  0°,  or  that  an 
angle  of  450°  is  equal  to  one  of  90°,  we  should  be  committing  the  same 
error. 

We  see  that,  while  an  angle  in  the  sense  of  elementary 
geometry  can  never  be  greater  than  180°,  our  new  concept 
of  angle  permits  us  to  speak  of  angles  of  any  magnitude. 

*  It  is  the  purpose  of  the  hour  hand  to  record  the  number  of  complete 
revolutions. 

133 


134     FUNCTIONS  OF  THE  GENERAL  ANGLE 

Our  notions  will  be  enriched  in  still  another  way,  if  we 
adopt  the  dynamic  instead  of  the  static  concept  of  angle. 
The  line,  whose  rotation  generates  the  angle,  may  revolve 
in  either  of  two  opposite  directions,  clockwise  or  counter- 
clockwise; and  we  must  distinguish  between  these  two  kinds 
of  rotation,  just  as  we  distinguish  between  two  motions  in 
opposite  directions  on  a  straight  line.  This  distinction  may 
be  made  by  ascribilig  to  every  angle,  not  merely  a  magnitude^ 
but  also  a  sign  depending  upon  the  direction  of  the  rotation 
by  which  the  angle  is  generated. 

It  18  customary/  to  speak  of  counterclockwise  rotations  as 
positive,  and  of  clockwise  rotations  as  negative. 

The  reason  for  this  convention  *  will  appear  later  (Art.  63). 

EXERCISE  XXXI 

/ 

Using  a  protractor,  combine  the  following  angles  graphically  and 
check  the  results  arithmetically. 

1.  30°  +  60°,  50°  -  30°,  30°  -  60°,  50°  +  (-  30°),  30°  +  (-  60°). 

2.  25°  +  15°  -  35°,  13.5°-  (-  25°)+  150°. 

3.  225°  +  345°  -  185°,  30°  +  (3  x  15°). 

4.  What  angle  does  the  minute  hand  of  a  clock  describe  in  3  hours 
and  25  minutes?  in  5  hours  and  13  minutes? 

5.  Suppose  that  the  dial  of  a  clock  is  transparent  so  that  it  may 
be  read  from  both  sides.  Each  of  two  persons,  stationed  on  opposite 
sides  of  the  dial,  observes  the  motion  of  the  minute  hand  for  fifteen 
minutes.  Upon  comparing  notes,  they  find  that  they  do  not  agree  in 
regard  to  the  angle  described  by  the  minute  hand  during  this  period  of 
time.     In  what  respect  do  they  differ? 

6.  What  is  the  magnitude  of  the  angle  described  by  a  spoke  of  a 
carriage  wheel,  3  feet  in  diameter,  when  the  carriage  travels  a  distance 
of  500  feet? 

Note.  Think  of  the  wheel  as  if  it  were  turning  on  the  axle  while  the 
carriage  is  standing  still. 

*  The  word  convention  is  here  used  in  a  special  sense,  meaning  an  arbitrary 
agreement. 


INITIAL   AND  TERMINAL   SIDE  135 

7.  The  earth  describes  an  approximately  circular  orbit  about  the 
sun  as  center  in  365  days.  What  angle  will  the  line  joining  the  sun  to 
the  earth  (the  earth's  radius  vector)  describe  in  415  days? 

8.  Two  wheels,  A  and  B,  are  joined  by  a  belt  as  in  Fig.  71.  The 
diameter  of  A  is  twice  that  of  B,  and  A  is  moving  in  counterclockwise 
direction.  What  angle  will  a  spoke  of 
B  describe  while  A  rotates  through  an 
angle  of  300°? 

9.  If  the  two  wheels  of  Ex.  8  are 

joined  by  a  crossed  belt,  what  angle  will 

a  spoke  of  B  describe  when  A   rotates 

through  an  angle  of  300'^  ? 

Fig.  71 
10.   If  n  wheels  are  connected  by  gears, 

what  kind  of  a  number  must  n  be  in  order  that  the  first  and  last  wheel 

may  rotate  in  the  same  direction?  in  opposite  directions? 

61.   Initial  and   terminal   side.      Standard  position   of   an 

angle.     Our  new  concept  of  an  angle,  as  a  measure  for  the 

/(7     amount  of  rotation  of  a  line,  leads  us 

y^  ^/       to  distinguish  between  the  initial  and 

o,        ^    p "^  terminal  sides  of  an  angle. 

■    Thus,  in  Fig.  72,  we  have  three  angles 
whose  senses  of  rotation  are   indicated  by 
\q^  curved  arrows.     The  angles  A  0\B  and  COzD 

Pjq  y2  *r®    positive,   for  the   rotation   is   counter- 

clockwise. Angle  EO3F  is  negative.  The 
initial  sides  of  these  angles  are  A  Oi,  CO2,  EO3,  and  their  terminal  sides 
are  BO^,  DO2,  and  FO3  respectively. 

If  an  angle  is  thought  of  as  generated  hy  the  rotation  of  a 
straight  line,  the  initial  and  final  positions  of  this  line  are  called 
the  initial  and  terminal  sides  of  the  angle  respectively. 

If  we  wish  to  compare  two  parallel  directed  line-segments 
in  regard  to  magnitude  and  sign,  we  usually  think  of  one  of 
them  as  being  moved,  until  its  initial  point  coincides  with  the 
initial  point  of  the  other.  In  the  same  way,  in  order  to  com- 
pare two  angles  we  usually  place  them  so  that  their  vertices 
and  initial  sides  shall  coincide. 

It  is  customary,  for  purposes  of  comparison,  to  place  all 


'^, 


136  FUNCTIONS  OF   THE   GENERAL   ANGLE 

angles  in  such  a  position  that  their 
initial  sides  are  on  a  horizontal  line 
and  pointed  toward  the  right.  An 
angle  placed  in  this  way  shall  be 
said  to  be  in  its  standard  position. 

Thus,  Fig.  73  represents  the  three  angles 

'^F  of  Fig.  72  in  their  standard  positions ;  the 

^^(^■'^^  angles  A  OB,  AOD,  and  AOF  oi  Fig.  73 

being  equal  to  the  three  angles  AO^B,   CO^D,  and  EO^F  of  Fig.  72 

respectively. 

EXERCISE  XXXII 

Place  the  following  angles  in  their  standard  positions  : 

1.  15°,  225°,  415^  768°. , 

2.  -  25°,  -  275°,  -  615°,  -  365°. 

3.  If  two  angles  differ  by  an  integral  multiple  of  360°  and  both 
angles  are  placed  in  standard  position,  how  will  the  terminal  sides  of 
the  two  angles  be  situated  with  respect  to  each  other  ? 

4.  If  two  angles,  placed  in  standard  position,  have  the  same  terminal 
side,  what  is  the  relation  between  them? 

5.  If  the  sum  of  two  angles  is  an  integral  multiple  of  360°,  how  will 
the  terminal  sides  of  the  two  angles  be  situated  with  respect  to  each 
other,  both  angles  being  placed  in  standard  position  ?       -^    J'<^  o 

62.  The  notion  of  the  trigonometric  functions  of  a  general 
angle.  Having  formulated  the  notion  of  a  general  angle^  it  be- 
comes necessary  to  revise  our  definitions  of  the  trigonometric 
functions,  since  our  original  definitions  are  applicable  to  acute 
angles  only.  To  be  sure,  we  have  already  made  some  prog- 
ress in  this  direction  by  defining  the  functions  of  an  obtuse 
angle.  (See  Arts.  40  and  42.)  But  those  definitions  were 
provisional,  and  it  will  be  advisable  to  reopen  the  whole  ques- 
tion, so  as  to  gain  a  larger  and  more  adequate  point  of  view. 

Our  new  concept  of  an  angle,  as  a  measure  for  the  amount 
of  rotation  of  a  line,  practically  forces  upon  us  the  following 
considerations  which  automatically  suggest  the  new  defi- 
nitions of  the  trigonometric  functions. 


RECTANGULAR  COORDINATES 


137 


Let  us  draw  a  positive  acute  angle  6  (see  the  Greek  alpha- 
bet on  page  132)  in  its  standard  position  (Fig.  74  a).  Let 
us  choose  a  point  P  anywhere  on  its  terminal  side  and  from 
P  drop  a  perpendicular  PM  to  the  initial  side.  Then,  in 
Accordance  with  the  definitions  of  the  functions  of  an  acute 
angle,  we  have 

MP  n      OM 


(1) 


sin 


e 


OP 


cos 


e 


OP 


Fig.  746 


Let  US  now  think  of  the  angle  0  as  growing.  Nothing 
remarkable  happens  until  6  reaches  90°.  At  that  moment, 
and  as  the  motion  continues,  our  original  definitions  cease  to 
be  applicable,  because  the 
right  triangle  POM,  of 
which  ^  is  an  interior  angle, 
ceases  to  exist.  But  we 
may  think  of  PMQ  (in 
Fig.  74  a)  as  a  plumb  line 
attached  to  a  point  P  on 
the  moving  terminal  side 
of  the  angle.  If  there  is  no  obstacle  at  0,  this  line,  re- 
maining always  vertical,  will  pass  from  the  right  to  the  left 
oi  0  2^  6  grows  from  an  acute  into  an  obtuse  angle  (see 
Fig.  74  5),  and  the  line-segment  OM  will  change  its  direc- 
tion. To  indicate  this  change  of  direction  of  OM,  we  rep- 
resent OM  by  a  positive  number  in  Fig.  74  a  and  by  a 
negative  number  in  Fig.  74  J.  If  we  count  the  distance  OP 
(which  does  not  change)  as  positive,  in  all  positions  of  the 
moving  line,  and  if  we  retain  equations  (1)  as  definitions 
for  sin  6  and  cos  6,  we  see  that  cos  6  becomes  negative  when 
0  becomes  obtuse.  The  line-segment  PM  does  not  change 
its  direction  until  6  grows  beyond  180°.  Therefore  the  sine 
of  an  obtuse  angle,  like  that  of  an  acute  angle,  is  positive. 
The  sine  of  an  angle,  however,  becomes  negative  when  the 
angle  lies  between  180°  and  360°. 


i 


63.  Rectangular  coordinates.     All  of  these  things  may  be 
stated  more  briefly  by  the  introduction  of  rectangular  coordi- 


138 


FUNCTIONS  OF  THE  GENERAL  ANGLE 


nates^  a  notion  of  utmost  importance,  not  merely  in  trigonom- 
etry, but  in  other  branches  of  mathematics. 

Let  us  draw  two  lines,  unbounded  in  length  and  perpen- 
dicular to  each  other.  We  shall  usually  think  of  one  of  them 
as  horizontal  and  call  it  the  a5-axis,  and  call  the  other,  which 
is  vertical,  the  ^/-axis.     The  point  0,  in  which  the  two  axes 

intersect,    is    called    the    origin   of 
coordinates. 

We  adopt  a  unit  of  length,  and 
denote  the  distances  from  any  point 
P  to  these  two  axes  by  x  and  y  re- 
spectively.    In  Fig.  75  we  have 

NP  =  Oiltf  =  X,  MP  =  0N=^  y, 

—^     where  the  notation  is  chosen  in  such 
a  way  that  x  is  measured  on  or  par- 
allel   to   the   2;-axis,  and   y  on   or 
parallel  to  the  y-axis. 
We  call  x  the  abscissa  and  y  the  ordinate  of  the  point  P. 
Both  numbers  together  are  called  the  coordinates  of  P. 

If  we  take  into  account  only  the  magnitudes  and  not  the 
directions  of  the  lines  OM^  OiV",  etc.,  that  is,  if  x  and  y  are 
regarded  as  numbers  without  sign,  there  will  be  four  points 
which  have  the  same 
coordinates. 

For  instance,  the  points  P, 
P',  Pi',  P'",  in  Fig.  76,  would 
all  correspond  to  a;  =  3,  y  =  2. 


M 


Fia.  75 


+ 

V 

p 

> 

+ 

2 

P 

+ 

1 

. 

3 

—  2 

-.1 

+1 

+2 

♦3 

0 

1 

pj 

2 

F'" 

In  order  to  avoid  this 
inconvenience,  we  intro- 
duce the  convention  that 
the  abscissas  of  all  points 
to  the  right  of  the  y-axis 
shall  be  positive,  and  of 
those  to  the  left  negative ;  that  the  ordinates  of  all 
points  above  the  rr-axis  shall  be  positive,  and  of  those  below 
negative. 


Fig.  76 


RECTANGULAR  COORDINATES  139 

The  coordinates  of  the  four  points  in  Fig.  76  are  now  different  from 

each  other. 

The  coordinates  of  P  are  a;  =  +  3,  3/  =  +  2, 
The  coordinates  of  P'  are  a;  =  —  3,  y  = -\-  2, 
The  coordinates  of  P"  are  a:  =  —  3,  y  =  —  2, 
The  coordinates  of  P'"  are  a:  =  +  3,  y  =-  2. 

The  positive  directions  of  the  x-  and  y-axes^  which  have  now 
been  defined,  will  hereafter  he  indicated  by  a  plus  sign  (as  in 
Fig.  76). 

The  distance  from  the  origin  0  to  any  point  P  is  usually 
denoted  by  r  and  is  called  the  radius  vector  of  that  point. 
We  shall  always  regard  the  radius  vector  as  positive^  and 
clearly  we  shall  always  have  (see  Fig.  77), 
r  =  4-  Va;2  -I-  y"^. 

Let  us  think  of  OP  (Fig.  77)  as  rotating  around  0  as  a 
center.  If  OP  originally  coincides  with  the  positive  a;-axis,  it 
will  require  a  counterclockwise  rota- 
tion of  90°,  or  a  clockwise  rotation  of 
270°,  to  bring  it  into  coincidence  with 
the  positive  2/-axis.  We  naturally  think 
of  the  numerically  smaller  angle  first, 
and  define  the  positive  sense  of  rotation 
to  be  that  one  which  enables  us  to  turn 
the  positive  a;-axis  into  the  position  of 
the  positive  ?/-axis  by  means  of  a  rota-  Fig.  77 

tion  of  only  90'^,     But  this  implies  that 

the  positive  direction  of  rotation   is   counterclockwise.     (Cf. 
Art.  60.) 

EXERCISE  XXXIII 

Plot  the  points  whose  coordinates  are  given  in  Exs.  1  to  5.  Find,  by 
measurement,  to  the  nearest  degree  for  the  angles  and  to  the  nearest  tenth 
of  a  unit  for  the  distances,  the  radius  vector  of  each  point  and  the  posi- 
tive angle  which  it  makes  with  the  positive  a:-axis.  Find  the  same 
results  by  calculation,  making  use  of  three-place  tables. 

1.  a:  =  +  3, 2/  =  +  4.  3.   a;  =  -  2.4,  y  = -\-  5.5. 

2.  ar  =  -t-  12,  y  =  -  5.  4.    a:  =  -  2.6,  y  =  -  2.1. 

5.   x  =  + 1.27,2/ =  -2.18. 


+  £0 


140      FUNCTIONS  OF  THE  GENERAL  ANGLE 

In  the  following  examples,  r  denotes  the  radius  vector  of  a  point  P, 
and  0  the  positive  angle  which  this  radius  vector  makes  with  the  posi- 
tive direction  of  the  a:-axis.  Plot  the  points.  Find,  by  measurement  to 
the  nearest  tenth  of  a  unit,  the  abscissas  and  ordinates  of  these  points. 
Find  the  same  results  by  calculation  with  three-place  tables. 

6.  r  =  2,e  =  30°.  8.    r  =  S,  0  =  210°. 

7.  r  =  o,e  =  135°.  9.   r  .=  4,  6  =  285°. 

10.   r  =  2.56,  ^  =  310°  20'. 

64.  Definition  of  the  trigonometric  functions  of  a  general 
angle.      We  are  now  in  a  position  to  give  the  definitions  of 

the  functions  of  a  general  angle  in 
a  compact  manner. 

Place  the  angle  6  in  its  standard 
position ;    that  is,  with  its  vertex 
on  the  origin  and  its  initial  side 
on  the  positive  a;-axis  of  a  system 
of  rectangular  coordinates.     (See 
+37     Fig.  78,  where  ^  is  a  positive  acute 
angle.)     Piek  out  a 'point  P^  differ- 
ent from   the    origin^  anywhere  on 
the  terminal  side  of  the  angle.      Then  we  adopt  the  folloiving 
definitions  : 

rr^,       .        £         ^     n  ordinate  of  P  -    a      1/ 

The  sine  of  angle  a         =  — r— ,  sm  0  =  - . 

radius  vector  oiP  ^ 

mi  '        £        ^    a  abscissa  of  P  a      « 

The  cosine  ot  angle  a     =  — -— ,         cos  0  =  -  . 

radius  vector  ot  P  ^ 

rr^i     ,         ^    £        1/1        ordinate  of  P  ^      a      1/ 

The  tangent  of  angle  &    =  -- — : r— -,  tan  0  =  ^» 

abscissa  oi  P  -^ 

The  cotangent  of  angle  6  =  — ^^— ^^ — — ,  cot  0  =  —  • 

ordinate  of  P  ^ 

rr«u  .    £        1/1         radius  vector  of  P  ^^^  a      r 

The  secant  ot  angle  u     = — : —— —  ,         sec  0  =  -  • 

abscissa  ot  P  **' 

rr.1  ^    £        1/1       radius  vector  of  P  ^^^  a      r 

The  cosecant  of  angle  6  — r-zr — ,         esc  0  =  —  • 

ordinate  of  P  *' 


EXCEPTIONAL  CASES  141 

As  Fig.  78  shows,  these  definitions  reduce  to  the  familiar 
definitions  of  Art.  7  if  ^  is  an  acute  angle.  In  the  case  of 
an  obtuse  angle,  they  give  results  which  agree  with  the  defi- 
nitions of  Arts.  40,  42,  and  43.  But  in  our  present  defini- 
tions, 6  is  not  restricted  either  in  magnitude  or  sign ;  it  may 
be  a  positive  or  negative  angle  of  any  magnitude.  The 
quantities  x  and  y  may  be  positive,  zero,  or  negative,  but  r 
is  always  positive. 

EXERCISE  XXXIV 

Construct  carefully  the  following  angles  and,  by  measurement,  find 
approximate  values  of  the  six  trigonometric  functions,  correct  to  two 
significant  places,  paying  particular  attention  to  their  signs. 

1.   25°.      2.   320°.      3.    110°.      4.    -  130°.      5.    +.725°.      6.    -  10°. 

7.  In  Art.  11,  the  exact  values  of  the  functions  of  30°,  45°,  and  60° 
were  expressed  by  means  of  radicals.  In  a  similar  way  find  the  values 
of  the  functions  of  the  following  angles : 

120°,  135°,  150°,  210°,  225,"  240°,  300°,  315°,  330°. 

8.  What  are  the  signs  of  the  trigonometric  functions  of  the  follow- 
ing angles : 

150°,  320°,   1000°,    -  625°. 

Find,  by  construction  and  measurement  to  the  nearest  degree,  the 
values  of  the  angles  for  which 

9.  >^^'  V    sin^  =  -t,   cos^  =  +  |.  >- 


.:(& 


10.      V  T/  tan  ^  =  1,      cosB  = 


1_ 

V2' 


11.  Show  that  the  values  of  the  trigonometric  functions  of  a  general 
angle,  as  given  by  the  definitions  of  Art.  64,  will  not  be  changed  if, 
instead  of  the  point  P,  any  other  point«»^'~6n  the  terminal  side  of  the 
angle  be  chosen.  /^  /u  .  /  \  :l 

65.  Discussion  of  the  exceptional  cases.  Each  of  the 
trigonometric  functions  is  defined  in  Art.  64  formally^  as  a 
quotient  of  two  numbers.  This  formal  definition  will  have 
a  real  significance  whenever  the  two  numbers  actually  have 
a  quotient.  Now  we  know  from  Algebra  that  two  numbers, 
B  (the  dividend)  and  d  (the  divisor),  always  have  a  unique 


142      FUNCTIONS  OF  THE  GENERAL  ANGLE 

quotient  q  if  the  divisor  is  different  from  zero.  That  is, 
there  exists  a  number  q  such  that 

or  what  amounts  to  the  same  thing,  such  that 

(2)  D  =  dq, 
whenever  d  is  different  from  zero. 

Now,  let  us  discuss  the  case  where  d  (the  divisor)  is  equal 
to  zero,  while  D  (the  dividend)  is  not.  In  this  case  there 
exists  no  number  q  which  satisfies  equation  (2).  For  this 
equation  now  becomes 

(3)  i>  =  0.^, 

and  its  right .  member  is  equal  to  zero  no  matter  what 
number  we  substitute  for  ^,  while  its  left  member  is,  b}^ 
hypothesis,  different  from  zero.  Consequently  it  involves 
a  contradiction  to  assume  that  a  number  has  been  obtained 
by  dividing  another  by  zero,  and  the  operation  of  dividing  by 
zero  is  therefore  excluded  from  Algebra. 

The  formal  definitions  of  Art.  64  involve  divisions  by  x^ 
y,  and  r,  and  therefore  lose  their  significance  in  any  case  in 
which  one  of  these  divisors  is  equal  to  zero.  Now  r=  OP 
is  the  radius  vector  of  a  point  P  which  may  be  chosen  any- 
where on  the  terminal  side  of  the  angle  except  at  0.  (Com- 
pare the  wording  of  the  definitions  in  Art.  64.)  Therefore 
r  is  never  equal  to  zero.     From  this  fact  and  the  equation 

(4)  a;2+?/2  =  r2, 

we  conclude  further  that  x  and  y  cannot  both  be  equal  to  zero 
at  the  same  time. 

It  will  now  be  clear  that  the  formal  definitions  of  Art.  64 
fail  to  provide  the  symbols 

...  tan  90°,      sec  90°,       tan  270°,      sec  270°, 

^  ^  cot  0°,        CSC  0°,  cot  180°,      CSC  180°, 

with  any  actual  meaning.  But  they  do  define  each  of  the 
six  trigonometric  functions  of  all  positive  angles  less  than 
360°  with  the  eight  exceptions  just  mentioned. 


EXCEPTIONAL  CASES  143 

If  the  angle  0  is  greater  than  360°,  or  if  it  is  negative, 
other  exceptional  cases  appear.  But  their  relation  to  the 
eight  exceptional  cases  (5)  is  so  simple  that  we  may  leave 
it  to  the  student  to  complete  this  discussion. 

Although  the  tangent  of  90°  is  not  defined,  our  definitions 
are  clearly  applicable  to  the  tangent  of  an  angle  6  which 
differs  from  90°  by  the  slightest  conceivable  amount.  Let 
us  see  how  the  tangent  of  an  angle  0  behaves  when  6  ap- 
proaches 90°  as  a  limit.  We  have, 
in  Fig.  79, 


+05 

Fig.  79 


where  P  may  be  any  point  different 

from  0  on  the  terminal  side  of  the 

angle.     For  our  present  purpose  it 

will    be    convenient   to   select   the 

point  P  in  the  following  manner.     Draw  a  line  RS  parallel 

to  the  rc-axis  at  any  convenient  distance  from  OM^  and  let 

P  be  the  intersection  of  the  terminal  side  of  the  angle  6 

with   this  fixed  line.     Then,  as  6  approaches  90°,  MP  =  y 

will  remain   constant,  while    OM=x  approaches  the  limit 

zero.     The  quotient  ^  =  tan  0  will  therefore  become  larger 

X 

and  larger.  Since  0M=  x  may  be  made  as  small  as  we 
please  by  taking  the  angle  6  =  MOP  close  enough  to  90°, 
while  the  ordinate  MP  always  remains  the  same,  we  see 
that  the  quotient  can  be  made  as  large  as  we  please.  In 
other  words,  the  angle  6  can  be  made  to  differ  so  little  from 
90°  that  its  tangent  will  become  larger  than  any  number 
whatsoever.  This  is  what  is  meant  by  the  statement  that 
tan  0  becomes  infinite  when  6  approaches  90°  as  a  limit.  We 
sometimes  express  this  same  statement  by  writing 

(6)  tan  90°  =00, 

a  symbolic  equation  which  should  be  interpreted  as  a  short- 
hand account  of  the  situation  which  has  just  been  described. 
It  is  not  a  definition  of  tan  90°.     For  oo  is  not  a  number,  and 


144      FUNCTIONS  OF  THE  GENERAL  ANGLE 

the  symbolic  equation  (6)  is  not  at  all  concerned  with  what 
happens  to  tan  6  when  6  is  equal  to  90°.  It  merely  tells  us, 
in  symbolic  form,  what  happens  when  6  approaches  90°  as 
a  limit ;  namely,  that  tan  6  then  increases  without  hound. 

In  the  preceding  discussion  we  considered  a  variable  angle 
MOP  which  approached  90°  as  a  limit.  In  order  to  see  what 
happened  to  its  tangent,  we  chose  the  point  P  on  the  terminal 
side  of  the  angle  in  such  a  way  that  its  distance  y  from  the 
a;-axis  remained  constant. 

We  may  obtain  the  same  result  in  a  slightly  different  way, 
which  may,  to  some  students,  appear  more  conclusive.  Let 
the  angle  MOP  approach  90°  as  before. 
Then  (Fig.  80), 

UnMOP  =  ^  =  ^. 
OM     X 

Let  us,  this  time,  choose  the  point  P  on 
the  terminal  side  of  the  angle  in  such  a 
■+£c  way  that  its  distance  x  from  the  y-axis 
remains  unchanged  while  d  approaches  90° 
as  a  limit.  It  is  evident  from  the  figure 
that  MP  =  y  will  then  increase  without  bound.  Therefore, 
we  obtain  again  the  result  that  tan  d  =  y/x  becomes  infinite 
when  6  approaches  90°  as  a  limit. 

So  far  we  have  tacitly  assumed  that  0  is  an  acute  angle 
increasing  toward  90°  as  a  limit.  What  happens  when  6 
starts  as  an  obtuse  angle  to  decrease  toward  90°  as  a  limit? 

Since  the  tangent  of  any  angle  between  90°  and  180°  is 
negative,  an  argument  precisely  similar  to  that  just  carried 
out  shows  that  the  numerical  value  of  tan  6  again  grows 
beyond  all  bounds  when  6  approaches  90°,  remaining,  how- 
ever, always  negative.  We  see  therefore  that  the  following 
statements  are  both  true  : 

1.  When  6  is  acute  and  increases  toward  90°  as  a  limit,  tan  6, 
remaining  always  positive^  grows  numerically  beyond  bound. 

2.  When  6  is  obtuse  and  decreases  toward  90°  as  a  limit, 
tan  ^,  remaining  always  negative^  grows  numerically  beyond 
bound. 


THE   FOUR   QUADRANTS 


145 


These  two  statements  are  frequently  summed  up  in  the 
symbolic  formula 
(7)  tan  90°  =  ±  00  . 

A  precisely  similar  discussion  will  show  that  tan  6  again 
becomes  infinite  when  6  approaches  270°,  that  sec  6  becomes 
infinite  when  0  approaches  90°  or  270°,  and  that  cot  6  and 
CSC  6  become  infinite  when  6  approaches  either  0°  or  180°. 
The  functions  sin  0  and  cos  6  are  always  finite. 

66.  The  four  quadrants.  The  x-  and  ^-axes  divide  the 
plane  into  four  portions  called  quadrants.  The  quadrant 
bounded  by  the  positive  x-  and  «/-axes  is  usually  called 
the  first  quadrant.  If  we  start  from  the  first  quadrant  and 
describe  a  path  around  the  origin  in  the  counterclockwise 
direction,  we  traverse  in  order  the  1st,  2d,  3d,  and  4th 
quadrants. 

An  angle  is  said  to  be  in  the  first,  second,  third,  or  fourth 
quadrant  according  to  the  quadrant  in  which  its  terminal 
side  falls  when  the  angle  is  in  its  standard  position,  that  is, 
with  its  initial  side  upon  the  positive  a;-axis. 

The  cardinal  angles  0°,  90°,  180°,  270°,  etc.,  may  be  regarded 
as  belonging  to  either  one  of  the  two  quadrants  upon  whose 
boundaries  they  lie. 

The  following  table  gives  the  signs  of  the  trigonometric 
functions  of  an  angle  in  the  various  quadrants  : 


I 

II 

III 

IV 

Sine 

+ 

+ 

- 

- 

Cosine    .... 

+ 

- 

- 

+ 

Tangent      .     .     . 

+ 

- 

+ 

- 

Cotangent  .     .     . 

+ 

- 

+ 

- 

Secant    .... 

+ 

- 

- 

+ 

Cosecant     .     .     . 

+ 

+ 

- 

- 

146 


FUNCTIONS  OF   THE   GENERAL  ANGLE 


EXERCISE    XXXV 

1.  Prove  each  of  the  following  symbolic  statements  and  explain  its 
significance  in  words. 

cotO°=±oo,    tan  90°  =  ±00,    cot  180°  =  ±  oo  ,    tan  270°  =  ±  cx> , 
csc0°=±oo,    sec  90°  =±00,    cscl80°  =  ±oo,    sec   270°  =  ±  oo  . 

2.  Show  that  the  numerical  value  of  the  sine  or  cosine  of  an  angle  can 
never  exceed  unity. 

3.  Show  that  I) / d  may  have  any  value  whatever  if  T>  and  d  are  both 
equal  to  zero,  and  hence  that  the  symbol  ^  is  wholly  indeterminate.  Why 
can  no  one  of  the  trigonometric  ratios  ever  have  this  form  ? 

4.  Determine  the  quadrants  of  the  following  angles  and  the  signs  of 
their  trigonometric  functions. 

325°,  710°,  1045°,  609°,  412°,  -  52°. 

5.  In  what  quadrant  is  an  angle  if  its  sine  and  cosine  are  both  posi- 
tive? If  its  sine  is  positive  and  its  tangent  negative  ?  If  its  secant  and 
tangent  are  both  positive  ? 

6.  If  we  know  that  the  sine  and  cosine  of  an  angle  have  the  same 
sign,  what  can  we  say  about  the  quadrant  of  the  angle  ? 

7.  Is  there  an  angle  whose  tangent  is  positive  and  whose  cotangent  is 
negative  ? 

8.  If  we  are  told  that  the  tangent  and  cotangent  of  an  angle  are  both 
positive,  does  this  enable  us  to  determine  the  quadrant  of  the  angle  ? 


67.     General  character  of  the  trigonometric  functions.    Their 
periodicity.      We  are  now  in  a  position  to  und^iiP^nd  liow  the 

functions  change  with  Mie  angle. 
For  the  purposes  of  this  discussion 
it  will  be  convenient  to  think  of  the 
radius  vector  r  as  constant.  This 
'  means  that  the  point  P,  which  ac- 
cording to  our  definition  must  be 
selected  on  the  terminal  side  of  the 
angle,  describes  the  circumference  of 
Fig.  81  a  circle  as  6  changes  from  0°  to  360°. 

(See  Fig.  81.)     It  is  easy  to  verify  the  following  statements 
by  reference  to  the  figure  : 


PERIODICITY  OF   FUNCTIONS 


147 


As  0  increases  from      0°  to    90°,  sin  6  increases  from  0  to  1. 
As  6  increases  from    90°  to  180°,  sin  6  decreases  from  1  to  0. 
As  6  increases  from  180°  to  270°,  sin  6  decreases  from  0  to  —1. 
As  6  increases  from  270°  to  360°,  sin  6  increases  from  —  1  to  0. 

It  is  also  evident  that  the  function  sin  6  repeats  its  values 
in  exactly  the  same  order  if  P  moves  around  the  circumference 
a  second,  third,  •••  nth.  time.  The  same  thing  is  true  of  the 
other  trigonometric  functions,  a  very  important  fact  which 
may  be  expressed  as  follows : 

Each  of  the  six  trigonometric  functions  is  periodic  and  its 
period  is  equal  to  360°.  That  is,  each  of  these  functions  repeats 
its  values  at  intervals  of  360°,  so  that 

sin  (O  +  n-  360°)  =  sin  6,   cos  (^  +  n  •  360°)  =  cos  0,  etc., 

where  n  is  any  positive  or  negative  integer  or  zero. 

The  behavior  of  each  of  the  six  functions  in  the  neighbor- 
hood of  each  of  the  four  cardinal  angles  may  be  recapitulated 
for  convenience  of  reference  in  the  following  table  : 


Sine 

COSINB 

Tangent 

Cotangent 

Secant 

Cosecant 

0° 

0 

+  1 

0 

TOO 

+  1 

Too 

'      90° 

+  1 

0 

±00 

0 

±00 

+  1 

180° 

0 

-1 

0 

Too 

-1 

±  00 

270° 

- 1 

0 

±20 

0 

Too 

- 1 

The  student  should  use  this  table  to  describe  in  words  the 
variation  of  each  of  the  six  functions  as  6  changes  from  0° 
to  360°. 

EXERCISE  XXXVI 

Discuss  the  variation  of  the  following  functions  as  0  varies  from  0°  to 
J60°. 

1.  cos^.  6.    sin  2^.  11.   sin  (- ^). 

2.  tan  e.  7.  2  sin  6.  12.  cos  {-6). 

3.  cot^.  8.  sin  3^.  13.  sin^^. 

4.  sec^.  9.  sin  4^.  14.  sin  ^  ^. 

5.  csc^.  10.  tan  4^.  15.  sin  (^  +  25°). 


148      FUNCTIONS  OF  THE  GENERAL  ANGLE 

68.  Relations  between  the  trigonometric  functions  of  a 
general  angle.  The  rel-ations  which  we  found  in  Art.  9, 
between  the  functions  of  an  acute  angle,  still  hold  without 
alteration  for  an  angle  of  any  magnitude.  In  fact  the  equa- 
tion between  the  abscissa,  ordinate,  and  radius  vector  of  a 
point  P,  that  is, ,  ^  _|_  ^2  _  ^.2^ 

is  true,  no  matter  in   what  quadrant  the   point  P  may  be 
situated. 

This  is  due  to  the  fact  that  only  the  squares  of  x,  y,  and  r  occur  in  this 
relation. 

If  wx  divide  both  members  of  the  above  equation  by  r^, 

we  find  (^^(y}\^i 

But,  by  the  definitions  of  Art.  64,  we  have,  in  all  quadrants, 

-  =  cos  6^,      ^  =  sin  (9, 
r  r 

so  that  the  preceding  equation  becomes 
(1)  sin2e  +  cos2e=l. 

Since  we  have,  by  definition. 


sin  (9  =  ^, 
r 

CSC  (9  =  -, 

y 

COS  ^  =  -, 
r 

sec  6'  =  -, 

X 

tan  l9  =  ^. 

cot  (9  =  -, 

X  y 

we  find  at  once 

(2)  sin  6  CSC  6  =  1,      cos  0  sec  0  =  1,      tan  0  cot  0  =  1. 

We  have  also    ,      ^      y  _  y/r  _  sin  6 

tan  u  —  —  —  — —  — -x-, 

X      x/r      cos  V 

which,  combined  with  (2),  gives  the  further  relations 

(3)  tan  0  = -,  cot  0  = -. 

^  ^  COS0  sin0 

If  we  divide  both  members  of  (1)  by  cos^  9  and  make  use 

of  (2)  and  (3),  we  find 

(4)  1+ tan2  0  =.sec2  0.     I 


TRIGONOMETRIC   IDENTITIES  149 

In  a  similar  fashion,  if  we  divide  both  members  of  (1)  by 
sin^  ^,  we  see  that 

(5)  1  +  cot2  e  =  csc2  e. 

When  the  angle  6  is  acute,  all  of  its  functions  are  positive. 
Consequently  if  oneTof  its  functions  is  given,  all  of  the  others 
may  be  found  without  ambiguity  by  means  of  the  above  rela- 
tions.    (Cf.  Art.  9  and  Exercise  VI,  Exs.  7-12.) 

But  if  we  do  not  know  in  what  quadrant  an  angle  lies  and 
are  given  the  value  of  merely  one  of  its  functions,  the  angle 
itself  and  its  other  functions  are  not  determined  uniquely. 

If  we  are  told,  for  instance,  that  sin  9  =  ^,  equation  (1)  only  tells  us 
that  cos2  ^  =  1  -  (J)2  =  3  . 

so  that  cos  B  =  ±l  V3, 

where  either  sign  may  be  taken.  In  fact  there  are  two  angles  between 
O''  and  360°  whose  sines  are  equal  to  J;  namely,  30°  and  150°.  We  may 
distinguish  between  them  by  stating  whether  the  cosine  is  positive  or 
negative. 

EXERCISE  XXXVII 

Find  the  other  functions  of  the  angle  0  as  determined  by  each  of  the 
following  conditions : 

1.  sin  ^  =  —  ^  and  ^  is  in  the  third  quadrant. 

2.  sin  ^  =  —  I  and  B  is  in  the  fourth  quadrant. 

3.  tan  B  =  +  2  and  B  is  in  the  third  quadrant. 

4.  cot  ^  =  —  3  and  sin  B  is  positive. 

5.  sec  ^  =  +  2  and  tan  B  is  negative. 

/'  6.  Find  the  values  of  the  other  functions  if  sin  B  =  a.  Are  all  pos- 
sible values  of  a  admissible  ?     State  a  reason  for  your  answer. 

7.  If  tan  B=  m,  find  the  values  of  the  other  functions. 

8.  If  sec  B  =  k,  find  the  values  of  the  other  functions.  Are  all  values 
of  k  admissible  in  this  problem  ?     State  a  reason  for  your  answer. 

69.  Trigonometric  identities  which  involve  functions  of  a 
single  angle.  By  means  of  the  relations  of  the  preceding 
article,  an  expression  which  involves  the  trigonometric  func- 
tions of  y ,  angle  0  may  be  written  in  a  great  many  different 
forms.  /^  ,1"^  often  important  to  be  able  to  recognize  that 
two  trigonometric  expressions,  although  different  in  form, 


150      FUNCTIONS  OF  THE  GENERAL  ANGLE 

are  really  identical.  This  may  frequently  be  done  by  inspec- 
tion. In  more  eomplicated  cases  it  is  advisable  to  express 
each  of  the  two  quantities,  whose  identity  we  wish  to  estab- 
lish, in  terms  of  some  one  of  the  six  functions  (the  sine,  for 
example).  It  will  then  become  evident  as  a  mere  matter  of 
algebra  whether  or  not  the  two  quantities  are  really 
identical. 

EXERCISE  XXXVIII 

1.  Show  that  sec  ^  —  tan  6  sin  6  —  cos  0,  for  all  values  of  6  for  which 
tan  6  and  sec  0  are  defined.     (See  p.  142.) 

Solution.     We  bave,  for  all  values  of  6,  for  which  tan  6  and  sec  6  are 
defined, 

a^^A      fo«^a;«/l         1         sin  ^^.    /J      1  — sin2^     cos^^  ^ 

sec  u  —  tan  a  sm  a  = ;: ;:  sin  u  = —  = ~  =  cos  u, 

cos  ^     cosp  cos^  cos  0 

which  proves  the  truth  of  the  original  assertion. 

2.  Prove  that  tan  ^  +  cot  ^4  =  sec  A  esc  A  is  an  identity.* 
Solution.     Denote  the  quantity  on  the  left  member  by  L  and  that  on 

the  right  member  by  R.     Then 

Z  =  tan^+cot^=5ia^  +  ^-2iii  =  «.iB!A±^2^= 1 , 

COS  A      sm  A         sin  A  cos  A         sin  A  cos  A 

R  =  sec  A  CSC  A  = 


cos  A  sin  A      sm  A  cos  A 
Therefore  L  =  R\  that  is, 

tan  A  +  cot  A  =  sec  ^  esc  A.  q.  e.  d. 

Prove  that  the  following  statements  are  identities : 

3.  cos  ^  tan  ^  =  sin  0.  7.   cos^^  -  sin^  A  =2  cos^  A  -  I. 

4.  sin  <^  cot  <^  =  cos  </>.  8-    (cse^  0  -  1)  sin^  0  =  cos^  $. 

J  5.   sin2  0  4-  sin2  $  tan2  $  =  tan2  $.   i^  9.    1  +  tan2  Q  = ?— — . 

1  —  sin2  d 
6.   cos2^-sin2^=l-2sin2^.      <io.   cos^  ^  — sin*  ^  =  2  cos2d  -  1. 

11.  (sin^  +  cos^)2+  (sin^-cos(9)2  =  2. 

12.  !^2l^2!ljZ_^^£ii^  =  csc^sec^. 

cos  ^  —  sin  ^ 

^3     cps  6>  cot  (9  -  sin  Q  tanjg^  i  +  sin  ^  cos  B. 
CSC  d  —  sec  ^ 

14.    J1-ZL!HL^  =  sec  ^  -  tan  6,  if  Q  is  an  acute  angle. 
^  1  +  sin  d 


*In  other  words,  show  that  the  left  member  is  equal  to  the  rig'^'^^member  for 
all  values  of  A  for  which  the  functions  tan  ^,  cot  ^,  sec  A,  c^^^ -^  have  been 
defined ;  that  is,  for  all  values  of  A  except  A  =  0°,  90°,  180°,  36( '^  etc. 


CHAPTER   X 

GRAPHIC   REPRESENTATIONS    OF    THE   TRIGONO- 
METRIC  FUNCTIONS 

70.   Line  representation  of  the  trigonometric  functions.     The 

trigonometric  functions  were  defined  as  ration  or  abstract 
numbers,  not  as  lines.  (See  Art.  7.)  This  does  not,  how- 
ever, preclude  the  possibility  of  representing  them  as  lines. 
An  abstract  or  concrete  quantity  of  any  hind  may  be  repre- 
sented as  a  line-segment,  by  choosing  arbitrarily  a  certain  line- 
segment  to  represent  a  unit  of  the  same  kind. 

Thus  we  may  represent  as  lines  the  populations  of  the  various  states 
of  the  Union,  taking  a  line-segment  one  inch  long  to  represent  a  popula- 
tion of  1,000,000.  The  populations  of  New  York  and^IUinois  will  then 
be  represented  by  line-segments  9.11  and  5.64  inches  in  length  respec- 
tively. Thus,  although  a  population  obviously  is  not  a  line-segment,  it 
may  be  represented  by  a  line-segment.  In  the  same  way  we  may  repre- 
sent the  values  of  the  trigonometric  functions  by  lines,  although  they  are 
not  lines,  but  abstract  numbers. 

The  following  is  a  convenient  method  for  obtaining  a 
representation  of  the  values  of  the  trigonometric  functions 
as  line-segments. 

We  construct  a  circle  with  the  origin  of  coordinates  as 
center.  An  angle  whose  vertex  is  at  the  center  of  this  circle 
will  subtend  an  arc  whose  numerical  measure,  in  degrees, 
minutes,  and  seconds,  is  equal  to  that  of  the  angle.  We 
may  therefore  speak  indifferently  either  about  the  functions 
of  the  angle  or  of  the  functions  of  the  arc.  The  point  in 
which  the  initial  side  of  the  angle  meets  the  circle  is  called 
the  origin  of  the  arc.  The  point  in  which  the  terminal  side 
of  the  angle  meets  the  circle  is  called  the  terminus,  or  the 
end  of  the  arc. 


152       GRAPHIC   REPRESENTATIONS   OF   FUNCTIONS 


If  an  angle  is  placed  in  its  standard  position,  the  origin  of 
the  subtended  arc  will  be  at  A,  the  point  in  which  the  positive 
a;-axis  meets  the  circle.  We  shall  call  this  point  the  primary 
origin  of  arcs.  The  point  B^  in  which  the  positive  ?/-axis 
intersects  the  circle  is  called  the  secondary  origin  of  arcs. 

Let  us  choose  any  convenient  unit  of  length,  say  an  inch, 
and  let  us  agree  to  measure  all  distances  in  terras  of  this 
unit.  We  then  construct  the  circle,  the  so-called  unit  circle^ 
whose  center  is  at  the  origin  of  coordinates  and  whose  radius 
is  equal  to  the  unit  of  length. 

In  Fig.  82,  \Qi  AOQ  =  6  he,  any  angle  in  its  standard  posi- 
j^y  tion,  and  AP  the  arc  which  it  sub- 

Is  tends  on  the  unit  circle.     Then 


(1) 


sm 


0^y_:=K^y^MP, 


r      1 


since  the  distance  r  =  OP  is  equal 
to  the  unit  of  length,  so  that  r  =  1. 
Now  X  =  OM  and  y  —  MP  are  the 
coordinates  of  P,  the  terminus  of  the  arc  AP.  Conse- 
quently we  may  express  our  result  as  follows  : 

If  any  angle  6  is  placed  in  its  standard  position^  the  value  of 
its  sine  is  equal.,  in  magnitude  and  sign.,  to  the  ordinate  of  the 
terminus  of  the  arc  ivhich  the  angle  subtends  on  the  unit  circle. 
The  value  of  its  cosine  is  equal  to  the  abscissa  of  the  terminus 
of  this  arc. 

In  order  to  find  a  line  representation  for  tan  6  and  cot  ^, 
we  draw  tangents  to  the  unit  circle  at  A  and  B  and  denote 
by  T  and  T'  the  points  in  which  the  terminal  side  of  the 
angle  intersects  these  two  tangents.     (See  Fig.  83.) 

Then  we  find 


tan.  =  ^=f=^2', 


cot^  =  tan^Or' 


BT'  ^  BT' 
OB        1 


=  BT', 


LINE   REPRESENTATION  OF   FUNCTIONS 


153 


since  the  radius  of  the  circle 

If  0  is  an  obtuse  angle,  OP  will  have  to  be  prolonged 
backward  in  order  to  intersect  the  tangent  at  A.  Moreover 
the    point   of  intersection 

T  will  then  be  below  A.  "  ,j,^Q 

Now  the  tangents  BT  and  --i^-^         I  ^-^^+it 

AT  are  parallel  to  the  x- 
and  i/-axes  respectively. 
Let  us  agree  to  give  signs 
to  the  line-segments  meas- 
ured on  these  two  tangents 
as  though  they  were  ab- 
scissas or  ordinates  of  a 
point.    That  is,  let  AT  \tQ 

positive  or  negative  according  as  T  is  above  or  below  A^ 
and  let  BT  be  positive  or  negative  according  as  T^  is  to 
the  right  or  left  of  B.  This  convention  is  indicated  in 
Fig.  83  by  the  two  -f-  signs  at  the  ends  of  the  two  tan- 
gents.    The  student  may  now  verify  that  the  equations 


Fig.  83 


(2) 


tan  ^=  at;   col  e  =  BT\ 


which  we  have  obtained  from  Fig.  83  in  the  case  of  an  acute 
angle,  will  give  correct  results  in  magnitude  and  sign^  no 
matter  in  what  quadrant  the  angle  0  may  happen  to  fall. 

We  may  formulate  our  results  as  follows : 

If  any  angle  6  is  placed  in  its  standard  position^  the  value  of 
its  tangent  is  equal,  in  magnitude  and  sign,  to  the  ordinate  of 
the  point  in  which  the  terminal  side  of  the  angle,  prolonged 
backward  if  necessary,  intersects  the  tangent  to  the  unit  circle  at 
the  primary  origin  of  arcs. 

The  cotayigent  of  the  angle  is  equal,  in  magnitude  and  sign, 
to  the  abscissa  of  the  point  in  which  the  terminal  side  of  the 
angle,  prolonged  backward  if  necessary,  intersects  the  tangent  to 
the  unit  circle  at  the  secondary  origin  of  arcs. 


154       GRAPHIC   REPRESENTATIONS  OF  FUNCTIONS 


(3) 


Referring  once  more  to  Fig.  83,  we  have 
a      OT      OT     rim 


OT'    or 


=  0T\ 


csGd  =  sGcBOr=  ^_ 

OjB         1 

These  line  representations  for  the  secant  and  cosecant  will 
hold,  in  magnitude  and  sign,  not  merely  for  acute  angles,  but 
for  angles  in  any  quadrant,  if  we  agree  to  make  the  following 
conventions  in  regard  to  sign.  OT  shall  be  positive  if  T  is 
on  the  same  side  of  0  as  P,  i.e.  if  ^  is  on  the  terminal  side 
of  the  angle  6.  OT  shall  be  negative  if  Tis  on  the  terminal 
side  of  the  angle  6  prolonged  backward.  OT'  shall  be 
positive  or  negative  according  as  T'  falls  on  the  terminal 
side  of  the  angle  6  or  on  the  terminal  side  prolonged  back- 
ward. 

We  leave  it  to  the  student  as  an  exercise  to  verify  these 
statements  in  detail  and  to  formulate  the  contents  of  equations 
(3)  in  words. 

Figures  84  to  87  illustrate  the  line  representation  of  the 


GRAPHS  OF   FUNCTIONS  155 

trigonometric  functions  for  an  angle  in  each  one  of  the  four 
quadrants.     In  each  of  these  figures 

MP  =  sin  e,     AT=  tan  <9,     0T=  sec  0, 

0M=  cos  e,    BT'  =  cot  6,    0T'  =  esc  0. 

These  line  representations  of  tau  6  and  sec  6  suffice  to  explain  why  the 
names  tangent  and  secant  were  chosen  for  these  functions.  The  word 
sine  is  not  capable  of  such  a  simple  explanation  and  has  a  long  and 
complicated  history. 

The  Greeks  did  not  use  the  six  functions  which  we 
have  introduced.  In  place  of  the  sine  of  an  angle  they 
made  use  of  the  chord  PQ,  subtended  by  the  angle  POQ, 
on  a  circle  of  known  radius.  (See  Fig.  88.)  If  the 
circle  has  a  unit  radius,  Fig.  88  shows  that  this  chord 
PQ  is  equal  to  twice  QR,  or  Fig.  88 

PQ  =  2  sin  ^  POQ. 

Thus  the  chord,  used  by  the  Greeks,  is  essentially  twice  the  sine  of 
half  the  angle. 

Aryabhata,  a  famous  Hindoo  mathematician  (born  476  a.d.),  was 
apparently  the  first  to  introduce  the  sine  of  the  angle  in  place  of  the 
chord,  and  he,  quite  naturally,  spoke  of  it  as  the  half-chord  or  jyd-ardhd, 
where  J  yd  is  the  Sanskrit  for  chord  or  bowstring  and  ardhd  for  one  half. 
For  the  sake  of  brevity  the  adjective  ardhd  was  soon  omitted  and  the 
sine  was  called  simply  jyd. 

The  Arabs,  who  far  more  than  any  other  people  cultivated  the  sciences 
during  the  Middle  Ages,  took  over  this  word  from  the  Hindoos,  but 
changed  its  spelling  to  jiba,  so  as  to  make  the  spelling  accord  with  the 
pronunciation  in  the  sense  of  their  own  language. 

But  in  written  Arabic  the  consonants  only  are  represented  by  definite 
characters,  the  vowels  being  merely  indicated  by  dots  which  are  fre- 
quently omitted  altogether.  As  a  consequence  of  this  practice,  the  Hin- 
doo word  jlba  was  soon  corrupted  into  jalb,  a  genuine  Arabic  word 
meaning  bosom,  heart,  or  pocket  according  to  the  context. 

In  the  twelfth  century,  when  the  Arabic  texts  were  translated  into 
Latin,  the  word  Jaib  was  translated  literally  by  the  Latin  word  sinus 
meaning  bosom. 

Thus,  a  foreign  word  was  first  converted  by  the  Arabs  into  a  word  of 
their  own  language  having  a  similar  sound  but  an  entirely  different 
meaning,  and  later.this  Arabic  word  was  translated  literally  into  Latin. 
Of  course  the  derivation  of  the  English  word  sine  from  sinus  is  obvious. 

71.   Graphs   of  functions,  a  number  of  whose  nixmerical 
ralues  are  given.      There  is  a  second  way  of  representing 


156       GRAPHIC   REPRESENTATIONS   OF   FUNCTIONS 


graphically  the  values  of  the  trigonometric  functions,  which 
is  even  more  important  than  that  which  has  just  been  dis- 
cussed. For  it  gives  us,  in  a  still  more  vivid  fashion,  a 
picture  of  all  the  most  essential  properties  of  these  functions  ; 
and  it  has  the  further  advantage  of  being  applicable,  not 
merely  to  the  trigonometric  functions,  but  to  all  of  the  other 
functions  which  naturally  arise  in  pure  and  applied  mathe- 
matics. In  order  tor  lead  the  student  to  appreciate  fully  the 
power  of  this  new  method,  we  shall  first  illustrate  it  by 
a  number  of  examples  taken  from  fields  other  than  trig- 
onometry. 


85  =  Date 

y  =  Population 

1790 

3,929,214 

1800 

5,308,483 

1810 

7,239,881 

1820 

9,638,453 

1830 

12,860,692 

1840 

17,063,353 

1850 

23,191,876 

1860 

31,443,321 

1870 

38,558,371 

1880 

50,155,783 

1890 

62,947,714 

1900 

75,994,575 

1910 

91,972,266 

1790  1800  1810  1820  1830  1840  1850  I860  1870  1880  1890  1900  1910 

Fig.  89 


The  population  of  the  United  States  is  determined  every  ten  years 
by  a  national  census.  The  table  in  the  margin  gives  the  results  of  these 
censuses.  It  is  customary  to  represent  the  contents  of  this  table  graph- 
ically by  laying  off  the  dates  horizontally  (i.e.  as  abscissas),  and  erect- 
ing for  each  of  these  dates  a  vertical  line  (ordinate),  which  shall  give 
by  its  length  in  terms  of  an  appropriately  chosen  unit  the  population  at 
that  time.  Figure  89  gives  such  a  graphic  representation  of  the  facts 
contained  in  this  population  table  and  presents  these  facts  in  a  more 
easily  intelligible  form  than  the  table  itself.  Moreover  if  we  join  the 
endpoints  of  the  ordinates  by  a  smooth  curve  (the  population  curve), 


GRAPHS  OF   FUNCTIONS 


157 


we  may  draw  some  fairly  reliable  conclusions  as  to  the  state  of  the 
population  in  the  years  1805,  1815,  etc.,  in  which  no  census  was  taken. 

If  we  plot  the  population  curves  of  two  or  more  countries  upon  the  same 
sheet,  a  great  many  interesting  matters  may  be  brought  out  by  comparison. 

Clearly  we  may  adopt  such  a  graphic  method,  whenever 
we  have  a  table  giving  a  relation  between  two  variables ; 
that  is,  a  table  which  shows  that  to  certain  numerical  values 
of  a  first  quantity  x  there  correspond  certain  numerical  values 
of  a  second  quantity  y. 

EXERCISE  XXXIX 
1.   The  following  table  gives  the  population  of  the  cities  of  New  York, 
Chicago,  and  Philadelphia  for  the  years  named : 


New  York  . 
Chicago  .  . 
Philadelphia 


1850 


515,547 

28,269 

340,045 


805,651 
109,206 
585,529 


1870 


942,292 
298,977 
674,022 


1,206,299 
503,298 
847,170 


1890 


1,515,301 
1,099,850 
1,046,964 


1900 


3,437,202 
1,698,572 
1,293,697 


4,766,883 
2,185,283 
1,549,008 


Make  a  graph  illustrating  this  information,*  and  from  the  graph  find 
the  probable  population  of  each  of  these  cities  in  1908. 

2.  Make  a  population  table  and  a  population  curve  for  the  city  and 
state  in  which  you  live. 

3.  Let  the  student  provide  himself  with  a  railroad  time-table,  giving 
the  names  of  the  various  stations,  their  distances  from  the  starting  point, 
and  the  times  at  which  a  certain  train  leaves  these  stations.  Draw  a 
distance-time  diagram  for  one  or  several  trains,  plotting  the  times  as 
abscissas  and  the  distances  as  ordinates. 

4.  On  April  3,  1912,  the  following  temperatures  were  observed  in 
Chicago : 

11  A.M. 

12  M. 

1  P.M. 

2  P.M. 

3  P.M. 

4  P.M. 

5  P.M. 

6  P.M. 


3  a.m. 

32° 

4  a.m. 

32° 

5  A.M. 

32° 

6  A.M. 

31° 

7  A.M. 

32° 

8  A.M. 

33° 

9  A.M. 

34° 

10  A.M. 

36° 

38° 

7  P.M. 

34° 

39° 

8  P.M. 

33° 

39° 

9  P.M. 

34° 

38° 

10  P.M. 

34° 

37° 

11  P.M. 

33° 

36° 

12  P.M. 

33° 

35° 

1  A.M. 

32° 

35° 

2  A.M. 

31° 

Represent  graphically. 


*  In  all  such  work,  involving  plotting  of  curves,  it  is  advisable  to  use  cross- 
section  paper. 


158       GRAPHIC   REPRESENTATIONS  OF  FUNCTIONS 

72.  Graphs  of  simple  algebraic  functions.  The  relation 
between  the  variables  x  aiid  ?/,  instead  of  being  given  by 
a  table  as  in  the  previous  examples,  may  be  given  by  an 
equation.  We  may  then  use  the  equation  for  the  purpose 
of  constructing  a  table,  and  then  draw  a  graph  as  before. 


Example  1.     Find  the  graph  oi  y  =  2x  —  b. 


Solution.     If   we   substitute   a:  =  0   in  the 
5 ;    for  a;  =  1,  we  find  y  - 


_  2 

-  1 

0 

+  1 

4-2 
+  3 
+  4 
+  5 


-9 

-7 
—  5 
-3 
-1 

+  1 
+  3 
+  5 


y 


given  equation,  we  find 
—  3  ;  etc.  We  construct 
in  this  way  the  table  printed  in  the  margin.  If  we  plot  the 
points  x=-2,  ?/=-9;  x--\,  y  =  -1\  etc.,  obtained 
in  this  way,  we  find  the  points  marked  in  Fig.  90  with  a 
little  cross.  All  of  these  points  are  found  to  be  on  a  straight 
line. 

This  observation  makes  it  seem  likely  that  all  of  the 
points  whose  coordinates  satisfy  the  equation  y  =^2x  —  b, 
not  merely  those  which  we  happened  to  compute,  are  on  this 
same  straight  line.  It  is  not  difficult  to  prove  that  this  is 
so,  but  the  proof  will  not  be  given  here. 


+x 


* 

/ 

, 

4-10          1 

T 

+  9              J 

\ 

1-8              / 

\ 

■♦■7             / 

\ 

1-6           / 

\ 

+5          / 

\ 

■♦•4       Y 

\ 

+3       / 

\ 

■•■2    / 

.  .  .  \: 

^7 

Fig.  90 


•  2-1    0    *■!  +2  t3+4 

Fig.  91 


+  a' 


Example  2.     Find  the  graph  oi  y  =  x^. 

Solution.     As  before,  we  construct  the  table  in  the  margin  by  comput- 
ing the  values  of  y  which  correspond  to  the  values  a:  =  —  3,  —  2,  —  1,  0, 


-2 

-1 

0 

+  1 
+  2 
+  3 


GRAPHS  OF   TRIGONOMETRIC   FUNCTIONS  159 

y  +  1,    +  2,   +  3.     We  then  plot  the   points  obtained  in   this 

—  way,  the  points  marked  with  a  little  cross  in  Fig.  91,  and 

^  unite  them  by  a  smooth  curve. 

4 

1         The  principle  involved  in  these  examples,  that  to 

0  every  equation  between  two  variables  x  and  y  there 
^  corresponds  a  curve  and  vice  versa,  is  at  the  founda- 
tion of  Analytic  G-eometry,  which  is  one  of  the  most 
important  developments  of  modern  mathematics. 
The  great  merit  of  having  introduced  this  idea  into 
mathematics  is  due  to  Descartes  (1596-1650)  and  Fermat 
(1601-1665). 

EXERCISE  XL 
Find  the  graphs  of  the  following  equations  : 
1.   y  =  x.  8.   y  =  2  r2  -  3  a;  +  1. 

2-   y  =  -x.  9.   y  =  x\ 

3.   y  =  -2x  +  \,  ^Q    ^^i^3_i. 

^.   y  =  2x\ 

5.  y  =  -x\  '   ^~^' 

6.  y  =  x"^  —  5.  1 

12.    y=±-l. 

7.  y  =  x^  —  5  X.  X 

73.  Graphs  of  the  trigonometric  functions.  We  now  pro- 
ceed to  apply  this  method  to  the  relation 

y  =  sin  X. 

We  choose  an  arbitrary  line-segment  on  the  a;-axis  to  repre- 
sent one  degree  and  another  arbitrary  line-segment  on  the 
?/-axis  to  represent  the  unit  value  of  the  sine,  that  is,  the 
abstract  number  1.  If  we  are  using  millimeter  paper,  or  a 
metric  scale,  it  will  be  convenient  to  make  a  distance  of  one 
millimeter  on  the  aj-axis  stand  for  one  degree,  and  to  measure 
the  ordinates  in  terms  of  a  unit  10  centimeters  or  100  mil- 
limeters long.  As  this  is  a  rather  large  scale  it  will  probably 
be  necessary  to  paste  several  sheets  together  in  order  to  be 
able  to  construct  the  whole  curve. 


160        GRAPHIC   REPRESENTATIONS   OF   FUNCTIONS 


From  the  table  of  natural  sines  we  obtain  the  table  in 
the  margin,  in  which  the  values  of  the  angle  x  as  well  as 
the  corresponding  values  of  sin  x  are  expressed  in  milli- 
metres in  accordance  with  the  adopted  scale,  which  makes 
1  mm.  on  the  a:-axis  stand  for  1°,  and  100  mm.  on  the  y- 
axis  stand  for  the  unit  value  of  the  sine.  This  table 
enables  us  to  plot  ten  points  of  our  curve  represent- 
ing ten  values  of  the  function  sin  x  in  the  first  quad- 
rant. (See  Fig.  92,  which  is  a  reduced  copy  of  such  a 
curve.) 

The  definition  of  the  sine  of  a  general  angle 

(Art.    64)    and  the  line    representation    of  the 

sine  in  the    unit   circle  (Art.   70),  both  show  very  clearly 

that  two  angles  like  80°  and  100°,  or  70°  and  110°,  which 


X 

y  =  smx 

0 

0 

10 

17 

20 

34 

30 

50 

40 

64 

50 

76 

60 

87 

70 

94 

80 

98 

90 

100 

Fig.  92.  —  The  Sine  Curve 

differ  by  the  same  amount  from  90°  but  in  opposite  direc- 
tions have  the  same  sine.  Consequently  that  portion  AB 
of  our  curve,  which  represents  the  values  of  sin  x  for  angles 
in  the  second  quadrant,  will  be  a  symmetric  counterpart  of 
the  first  portion  OA^  which  corresponds  to  angles  in  the 
first  quadrant.     (See  Fig.  92.) 

The  unit  circle  also  makes  it  evident  that  the  sines  of  two 
angles  which  differ  by  180°  are  numerically  equal  .but 
opposite  in  sign.  Consequently  that  portion  BCD  of  our 
curve,  which  corresponds  to  angles  in  the  third  and  fourth 
quadrants  and  all  of  whose  ordinates  are  negative,  may  be 
obtained  easily  from  the  known  part  OAB.     The  parts  OAB 


GRAPHS   OF   TRIGONOMETRIC    FUNCTIONS 


161 


and  BCD  of  the  curve  are  in  fact  congruent,  but  are  situated 
on  opposite  sides  of  the  a;-axis. 

We  have  already  noted  that  the  sine  function  repeats  its 
values  at  intervals  of  360°.  It  is  a  periodic  function  with  a 
period  of  360°  (Art.  67).     This  manifests  itself  in  the  graph 


+  1/ 


+  ar 


Fio.  93.  —  The  Cosine  Curve 

by  the  fact  that  the  picture  in  each  of  the  intervals  from 
360°  to  720°,  etc.,  from  -  360°  to  0°,  etc.,  is  an  exact  copy 
of  that  piece  of  the  curve  which  lies  between  0°  and  360°. 

The  curve  obtained  in  this  way  from  the  sine  function  is 
called  the  sine  curve. 

The  cosine  curve,  which  is  the  graph  of  the  function 

y  —  cos  2;, 

is  of  the  same  general  character  as  the  sine  curve.  Its  form 
is  given  in  Fig.  93,  and  may  be  obtained  by  applying  to  the 
cosine  function  an  argument  exactly  similar  to  that  which 
has  just  been  carried  out  for  the  sine. 

We  may  apply  the  same  method  to  the  function 

y  =  tan  x. 

However,  the  graph  obtained  in  this  way  differs  very  essen- 
tially from  the  sine  and  cosine  curves. 

In  fact  we  know  that  when  x  approaches  90°  from  below, 
that  is,  if  X  assumes  a  succession  of  values  like 

89°,  89°.9,  89°.99,  89°.999,  etc., 


162       GRAPHIC   REPRESENTATIONS   OF   FUNCTIONS 

the  tangent  of  a;,  remaining  always  positive,  will  grow  numeri- 
cally beyond  all  bound.  If  on  the  other  hand  x  approaches 
90°  from  above,  through  a  sequence  of  values  like 

91°,  90°.l,  90°.01,  90°.001,  etc., 

-I 
the  tangent  of  ic,  remaining  always  negative,  again  grows 

numerically  beyond  all  bound.     (Cf.   Art.   Qb.^     We  see, 

therefore,  that  the  difference  between  the  values  of 

tan  (90°  +  K)  and  tan  (90°  -  7i) 

grows  larger  and  larger  as  the  angles 

90°  +  h  and  90°  -  h 

themselves  come  closer  and  closer  together. 

We  express  this  by  saying  that  the  function  tan  x  is  discon- 
tinuous for  X  =  90°.  The  corresponding  property  of  the 
graph  is  an  interruption  or  break  in  the  otherwise  continuous 
curve.  There  are  no  such  breaks  in  the  sine  or  cosine  curves. 
We  can  think  of  a  material  point  (say  the  point  of  a  lead 
pencil)  as  actually  describing  a  sine  curve  without  interrup- 
tion. If  we  were  to  attempt  to  do  the  same  thing  for  the 
tangent  curve,  we  should  have  to  interrupt  the  path  of  the 
point  at  a;  =  90°,  at  a;  =  270°,  etc. 

We  meet  here  the  important  distinction  between  continuous 
and  discontinuous  functions,  the  precise  formulation  of  which 
must  be  left  to  a  later  point  in  the  student's  career. 

The  tangent  is,  of  course,  a  periodic  function  and  repeats 
its  values  at  intervals  of  360°.  But  we  may  now  observe 
that,  unlike  the  sine  or  cosine,  it  repeats  its  values  after  the 
shorter  interval  of  180°.  To  recapitulate:  the  tangent  is  a 
periodic  function  of  period  180°,  and  is  discontinuous  for 
X  =  90°  and  for  all  values  of  x  which  differ  from  90°  b^  integral 
multiples  of  180°. 

Figure  94  shows  the  form  of  the  tangent  curve. 

EXERCISE  XLI 

1.  Plot  the  curves  y  =  2  sin  x,  y  =  dsinx,  y  =  4  sin  x. 

2.  Plot  the  curves  y  =  8m2  x,  y  =  sin  3  ar,  y  =  sin  4  x. 


RADIAN   MEASURE 


163 


3.  How  are   the  curves   of   Exs.   1   and   2    related    to    the    curve 
2^  =  sin  X  ? 

4.  Show  that  the  curve  y  =  cot  x  is  discontinuous  for  x  =  0°,  180°, 
360°,  etc.,  and  has  the  form  indicated  in  Fig.  95. 

+y 


/ 


Fig.  94.  —The  Tangent  Curve 


Fig.  95.  —  The  Cotangent  Curve 


5.   Show  that  the  curves  ?/  =  sec  x  and  y  =  esc  a;  have  the  forms  indi- 
cated in  Figs.  96  and  97. 


0°       3C0° 


Fig.  96.  — The  Secant  Curve 


270°     3C 


90"       IS 


Fig.  97.  —The  Cosecant  Curve 


L 


74.  The  natural  unit  of  circular  measurement.  Definition 
of  a  radian.  In  constructing  the  graphs  of  the  trigonometric 
functions,  the  student  may  have  observed  that  the  units  of 
measurement  on  the  x-  and  z/-axes  were  both  chosen  arbi- 
trarily, and  might  have  been  selected  in  infinitely  many 
different  ways,  thus  altering  materially  the  appearance  of 
the  resulting  curve. 

This  mutual  independence  of  the  two  scales,  on  the  two 
coordinate  axes,  is  a  natural  consequence  of  the  fact  that 


164       GKAPHIC   REPRESENTATIONS  OF  FUNCTIONS 

the  two  quantities  x  and  y  were  regarded  as  different  in 
kind.  One  of  them  was  regarded  as  an  angle  measured  in 
degrees,  and  the  other  as  an  abstract  number.  Having 
chosen  a  certain  horizontal  line-segment  as  representative 
of  the  unit  of  angles  (1°),  it  was  still  admissible  to  choose 
arbitrarily  another  (vertical)  line-segment  to  represent  the 
unit  of  abstract  numbers.  Whenever  x  and  y  represent  two 
quantities  of  different  kind,  the  x-  and  y-scales  are,  in  the 
nature  of  things,  independent  of  each  other. 

Even  if  x  and  y  are  quantities  of  the  same  kind,  it  is  often 
more  convenient  to  choose  the  lengths  of  the  units  different 
on  the  two  scales.  If  this  were  not  done,  the  resulting 
curve  might  fail  utterly  to  serve  the  purposes  for  which  it 
was  intended. 

Thus,  when  we  draw  a  profile  map  of  an  extensive  country  (showing 
the  elevations  of  various  points  in  a  certain  vertical  cross  section),  the 
vertical  scale  must  be  chosen  much  larger  than  the  horizontal  scale. 
Otherwise  the  differences  of  elevation,  as  depicted  on  the  map,  would 
become  so  small  as  to  be  unnoticeable. 

Nevertheless  it  will  usually  be  desirable  to  choose  the  hori- 
zontal and  vertical  units  equal  to  each  other  whenever  x  and 
y  may  be  regarded  as  quantities  of  the  same  kind,  provided 
that  the  resulting  curve  does  not  thereby  lose  its  usefulness 
as  it  would  in  the  example  just  quoted. 

Now  the  considerations  of  Art.  70  show  that,  from  a  cer- 
tain point  of  view,  the  quantities  x  and  y  which  occur  in 

such  an  equation  as  .,  _    -^  ^ 

»  y  —  Sill  X 

may  be  regarded  as  quantities  of  the  same  kind. 

In  fact,  we  observed  in  Art.  70  that  we  might  think  of 
the  number  x  as  the  measure  of  the  arc  AP  (Fig.  82)  in- 
stead of  as  the  measure  of  the  corresponding  angle  AG  P. 
We  saw  further  that  certain  line-segments  could  be  con- 
structed whose  lengths,  in  terms  of  the  radius  of  the  circle 
as  unit,  were  equal  to  the  values  of  sin  x^  cos  x^  etc.  If  then 
we  measure  the  length  of  the  arc  x  in  terms  of  the  radius 
of  the  circle  as  ninit,  instead  of  in  degrees,  we  shall  have  the 


I 


RADIAN  MEASURE  166 

arc  and  its  trigonometric  functions  expressed  in  terms  of  the 
same  unit. 

This  unit  of  arc  measure^  an  arc  of  a  circle  whose  length  is 
equal  to  the  radius  of  the  circle^  is  called  a  radian. 

One  advantage  gained  by  measuring  arcs  in  radians  is 
this :  the  arc  and  its  trigonometric  functions  will  then  be 
expressed  in  terms  of  the  same  unit. 

If  r  is  the  radius  of  a  circle,  the  length  of  its  circumfer- 
ence is  equal  to  2  7rr.  Therefore  a  circumference  may  be 
said  to  contain  2  tt  radians.  Since  it  also  contains  360  de- 
grees, we  have 

(1)  2  TT  radians  =  360°, 

whence 

1       ,.          360°      180° 
1  radian  =  — —  = . 

2  TT  TT 

Since  tt  =  3.14159265,  we  find,  to  seven  decimal  places, 

(2)  ^       1  radian  =  57°.2957795. 
On  the  other  hand  we  find  from  (1)  •    ' 

(3)  l°  =  j|^  radians, 
or 

(4)  1°  =  0.0174533  radian. 

These  equations  make  it  easy  to  find  the  number  of  de- 
grees in  an  angle  or  arc  when  its  measure  is  given  in  radians, 
or  vice  versa. 

Clearly  it  follows,  from  the  definition  of  a  radian,  that  the 
length  of  the  arc  which  an  angle  of  one  radian  intercepts  on 
the  circumference  of  a  circle  of  radius  r  is  itself  equal  to  r. 
Then,  an  angle  of  half  a  radian  at  the  center  will  intercept 
an  arc  on  the  circumference  whose  length  is  equal  to  ^r. 

In  general :  an  angle  of  6  radians  at  the  center  of  a  circle  of 
radius  r  intercepts  an  arc  s  upon  the  circumference  whose  length 
is 

(5)  '  s  =  re. 


166      GRAPHIC  representatio:ns  of  functions 

In  the  simplicity  of  this  formula  lies  a  second  great  ad- 
vantage of  measuring  angles  in  radians  rather  than  in  de- 
grees. 

EXERCISE    XLII 

Convert  into  radians  the  following  angles: 

~L.   90°.  /^.   30°.  5.    +693°  20'. 

^2.   270°.  4.    +25°  15'.  6.    -  1030°  0'. 

Convert  into  degrees  the  following  angles  which  are  given  in  radians, 
state  their  quadrants  and  the  signs  of  their  trigonometric  functions : 

>^.    -.  (^9.   ^.  11.   0.7691. 

2  16 

l«.  E.  10.   3.14159.  12.   5.3214. 

4 

1^    13.    Prove  that  the  area  of  a  sector  of  a  circle  of  radius  r  is  equal  to 
\  r^O  if  0,  the  angle  at  the  center,  is  measured  in  radians. 

14.  Prove  that  a  segment  of  a  circle  of  radius  r,  whose  arc  is  equal 
to  6  radians,  has  the  area 

^r2(^-sin^). 

15.  Compute  the  area  of  a  circular  segment  of  radius  11  feet,  if  its 
arc  is  equal  to  52°. 

16.  How  will  the  formulae  of  Exs.  13  and  14  be  modified,  if  the  angle 
6  is  expressed  in  degrees  ? 

17.  A  cord  is  stretched  around  two  wheels,  a  large  one  of  radius  r 
and  a  smaller  one  of  radius  r'  feet,  the  distance  between  the  centers  of 
the  wheels  being  d  feet.  If  the  cord  is  not  crossed  and  if  ^  is  the  angle 
of  inclination,  expressed  in  radians,  of  the  free  part  of  the  cord  to  the 
line  of  centers  of  the  wheels,  show  that 

sin  0  =  '-=^,    / .;  2  p  cos  ^  +  /'I  +  ^) r  +  (l  -  ^^'J, 

where  I  is  the  entire  length  of  the  cord. 

18.  If  the  cord  in  Ex.  17  is  crossed,  show  that  its  length  I  may  be 
found  by  means  of  the  formulae 


6=1+^,1 

d 


2[rfcos^  +  (|+^](r  +  r')]- 


19.  Find  the  length  of  a  belt  which  is  to  be  stretched  around  two 
wheels  3  and  2  feet  in  diameter  respectively,  if  the  distance  between  the 
centers  of  the  two  wheels  is  5  feet :  (a)  if  the  belt  is  crossed,  (b)  if  it  is 
not  crossed. 


FUNCTIONS  OF   SYMMETRICAL  ANGLES 


167 


20.  Draw  the  graphs  of  the  trigonometric  functions  sin  a:,  cos  x,  tan  x, 
cotar,  if  x  is  expressed  in  radians,  using  the  same  unit  of  length  for  x 
distances  and  y  distances. 


75.  Relations  between  the  functions  of  two  symmetrical 
angles.  Let  us  consider  two  angles,  like  90°  —  ^  and  90° -f^, 
which  differ  from  one  of  the  cardinal  angles  by  the  same 
amount  but  in  opposite  directions.  If  we  place  two  such 
angles  in  the  standard  position,  their  terminal  sides  will  be 
symmetrically  situated  with  respect  to  one  of  the  two 
coordinate  axes,  so  that  this  axis  will  bisect  the  angle  be- 
tween them.  As  a  consequence  of  this  fact  the  trigo- 
nometric functions  of  the  two  angles  are  related  to  each 
other  in  a  very  simple  fashion. 

In  order  to  obtain  these  relations  we  shall  consider  each 
of  the  four  cardinal  angles  separately,  making  use  of  the 
line  representation  of  the  functions 
given  in  Art.  70. 

Let  us  begin  with  the  cardinal 
angle  0°  or  0  radians.  Figure  98 
represents  the  unit  circle  and  the 
two  angles 

^  =  ZAOPand  -  6  =  Z.  AOP\ 

each  in  its  standard  position.     The 

arcs    AP  '  and   AP'    subtended   by  Fio.  gg 

these  angles  on  the  unit  circle  are 

symmetrical  with  respect  to  A.     Therefore   the  ordinates 

of  P  and  P'  (the  termini  of  these  arcs)  are  numerically 

equal  and  opposite  in  sign,  while  their  abscissas  are  equal  in 

magnitude  and  sign. 

But  the  ordinate  and  abscissa  of  the  terminus  of  the  arc 
AP  are  respectively  equal  to  the  sine  and  cosine  of  6, 
The  ordinate  and  abscissa  of  P'  are  respectively  equal  to  the 
sine  and  cosine  of  —  6.     (See  Art.  70.) 

mk     Consequently  we  have 

^Kl)  sin  (  —  ^)  =  —  sin  ^,  cos  (  —  ^)  =  cos  6, 


168        GRAPHIC   REPRESENTATIONS  OF   FUNCTIONS 


Consider  next  the  case  of  two  angles  90°  —  6  and  90°  +  B 
symmetric  with  respect  to  the  cardinal  angle  90°.     In  this 

case  (see  Fig.  99)  P  and  P'  have 
the  same  ordinate,  while  their  ab- 
scissas are  numerically  equal  but 
opposite  in  sign.     Consequently 

sin  (90°  -  ^)  =  sin  (90°  +  (9), 


4 

y 

A 

^\ 

I        ^ 

r 

(2  a) 


cos(90°-^)  =  -cos(90°  +  6>), 


Fig.  99 


or,   if   the   angles   are   measured   in 
radians, 

(2  5)      sing  -  ^)=  sin(|  +  6^,  cos(|  -  6^=  -  cos(|  +  ^). 

By  a  precisely  similar  argument,  the  details  of  which  we 
leave  to  the  student,  we  find 


(8  a) 
or, 
(3  6) 


sin  (180°  -6)=-  sin  (180°  +  (9), 
cos  (180°  -  (9)  =  cos  (180°  +  (9)  ; 

sin  (tt  —  ^)  =  —  sin  (tt  +  0), 
cos(7r  —  ^)  ==  cos  (tt  +  ^), 


according  as  the  angle  is  measured  in  degrees  or  radians, 
and  also 


(4  a) 

or, 

(4  5) 


sin  (270°  -0)=  sin  (270°  +  <9), 
cos  (270°  -6)=-^  cos(270°  +  (9)  ; 

.„(^-.)..i.f^.«). 


Of  course  we  shall  also  have 

sin  (360°  -  ^)  =  -  sin  (360°  +  (9), 
cos  (360°  -  <9)  =  cos  (360°  -f  Oy 


I 


RELATIONS  AMONG  THE   FUNCTIONS  169 

But  these  equations  are  really  repetitions  of  (1)  if  we 
remember  that,  on  account  of  the  periodicity  of  the  sine 
and  cosine, 

sin  (360°  +  ^)  =  sin  <9,  sin  (360°  -  0)  =  sin  (  -  (9), 
cos  (360°  +  6)=  cos  6,  cos  (360°  -  (9)  =  cos  (-  (9). 

The  relations,  which  correspond  to  (1),  (2),  (3),  (4)  for  the 
remaining  trigonometric  functions,  may  easily  be  obtained 
by  expressing  the  tangent,  cotangent,  secant,  and  cosecant  in 
terms  of  the  sine  and  cosine.  (See  Art.  68.)  Thus,  for 
instance, 

tan(-  0)^  ^^"^-  ^)  =  -  '^'f=-  tan  0. 
^       ^     cos(-^)        cos^ 

The  student  should  actually  work  out  the  sixteen  equations 
obtainable  in  this  way  and  combine  them  in  tabular  form 
with  the  eight  equations  (1)  to  (4). 

'The  student  should  also  observe  that,  although  we  have 
constructed  the  figures  for  the  case  when  ^  is  a  positive  acute 
angle,  our  proof  of  formulae  (1),  (2),  (3),  (4)  remains  valid 
word  for  word,  if  ^  is  a  positive  or  negative  angle  of  any 
magnitude.  A  good  way  to  convince  one's  self  of  this  fact  is 
to  think  of  the  angle  0  as  variable  and  to  follow  out  mentally 
the  changes  which  would  take  place  in  such  a  figure  as 
Fig.  98  or  Fig.  99  when  the  angle  6  increases  or  decreases. 
The  fact  that  equations  (1)  to  (4)  are  universally  valid  will 
thus  be  rendered  intuitive. 

76.  Relations  between  functions  of  two  angles  whose  sum 
or  difference  is  a  right  angle.  If  0  is  an  acute  angle,  we 
know,  from  Art.  10,  that " 

(1)  sin  (90°  -6)=  cos  0,  cos  (90°  -  d)=  sin  0. 

If  we  combine  these  equations  with  (2  a)  of  Art.  75,  we  find 
further 

(2)  sin  (90°  +  6}=  cos  (9,  cos  (90°  +  0}  =  -  sin  6. 

We  wish  to  show  that  the  four  equations  (1)  and  (2)  are 
true,  not  merely  when  6  is  an  acute  angle,  but  when  ^  is  a 


170       GRAPHIC   REPRESENTATIONS   OF   FUNCTIONS 

positive  or  negative  angle  of  any  magnitude.  This  may  be 
done  by  the  method  of  mathematical  induction. 

We  begin  by  proving  the  following  theorem.  If  equations 
(1)  and  (2)  are  true  for  a  certain  angle  0,  they  are  also  true 
for  the  angle  d'  =  90°  +  0. 

Proof.  By  hypothesis,  equations  (1)  and  (2)  are  true 
for  the  angle  6.     Therefore  we  have 

sin  6'  =  sin  (90°  +  <?)  =  cos  (9,  cos  6'  =  cos  (90°  +  (9)  =  -  sin  6. 
But 

sin  (90°  -  ^0  =  sin  [90°  -  (90°  +  (9)]  =  sin(-  6>)  =  -  sin  d, 
and 

cos  (90°  -  (90  =  cos  [90°  -  (90°  +  ^)]  =  cos  (-  l9)  =  cos  6, 

(Art.  75,  equations  (1)), 
which  proves  that 

(3)  sin  (90° -^0=  cos  (9^  cos  (90°  -  6")  =  sin  (9', 

since  both  members  of  the  first  equation  are  equal  to  —  sin  6^ 
and  both  members  of  the  second  are  equal  to  cos  6. 

Since  equations  (2  a)  of  Art.  75  are  true  for  all  angles,  we 
now  find 

sin  (90°  +  6'^  =  sin  (90°  -6')=  cos  6', 
^  ^         'cos  (90°  +  (90  =  -  cos  (90°  -0')=-  sin  (9^ 

Since  equations  (3)  and  (4)  are  the  same  as  (1)  and  (2), 
with  6'  in  place  of  ^,  we  have  actually  proved  our  theorem; 
namely,  if  equations  (1)  and  (2)  are  true  for  the  angle  ^, 
they  are  also  true  for  the  angle  6'  =  90°  +  0. 

We  know  that  equations  (1)  and  (2)  are  true  for  all  posi- 
tive acute  angles.  As  a  consequence  of  the  theorem  just 
proved,  they  are  successively  seen  to  be  true  for  all  positive 
angles  in  the  second,  third,  or  fourth  quadrant,  and  conse- 
quently for  all  positive  angles  whatever. 

But  they  are  also  true  for  all  negative  angles.  For  let  0 
be  a  negative  angle.  Let  n  •  360°,  where  w  is  a  positive  in- 
teger, be  the  lowest  integral  multiple  of  360°  which  makes 

0'  =  e  +  n'  360° 


THE   QUADRANTAL   FORMULAE  171 

a  positive  angle.     Then,  on  account  of  the  periodic  character 
of  the  sine  and  cosine,  we  shall  have 

(5)  sin  e'  =  sin  (l9  +  w  •  360°)=  sin  6, 

cos  d'  =  cosle  4-  n  '  360°)  =  cos  (9, 

and  similarly 

sin(90°-6>')  =  sin(90°-^),  cos(90°-^')  =  cos(90°-^), 
^^    sin(90°  +  l9')  =  sin(90°+(9),  cos(90°  +  l9')  =  cos(90°  +  6'). 

Since  0'  is  a  positive  angle,  we  have 

sin  (90°  -  ^0  =  cos  l9^  cos  (90°  -  ^')  =  sin  0', 
sin  (90°  +  (90  =  cos  (9',  cos  (90°  +  (9')  =  -  sin  6^. 

If  in  these  equations  we  substitute  the  values  (5)  and  (6), 
we  find 

sin  (90°  -  <9)  =  cos  (9,  cos  (90°  -  <9)  =  sin  ^, 
sin  (90°  +  ^)  =  cos  ^,  cos  (90°  +  ^)  =  -  sin  ^. 

Therefore  e^'wa^zo /IS  (1)  and  (2)  are  true  for  positive  and  nega- 
tive angles  of  any  magnitude. 

The  formulae  for  tan  (90°  +  ^),  sec  (90°  +  (9),  etc.,  may  be 
found  by  expressing  these  functions  of  90°  +  ^  in  terms  of 
sin  (90°  +  6)  and  cos  (90°  +  6)  and  making  use  of  (2) ;  for 
instance,  we  find 

tan  (90°  +  ^)  =  ^^^CQQ°  +  f>  =  -^^  =  -  cotk 
^  ^      cos (90° +  (9)      -sin  (9 

77.  The  quadrantal  formulae.  If  we  unite  equations  (1)  of 
Art.  76  with  the  corresponding  formulae  for  the  remaining 
four  functions,  we  obtain  the  following  system  of  equations: 

sin  (90°  -  <9)  =  cos  0,         cos  (90°  -  (9)  =  sin  d, 

(1)  tan  (90°  -  <9j)  =  cot  d,         cot  (90°  -  ^)  =  tan  (9, 
sec  (90°  -6)=  CSC  d,         esc  (90°  -  6>)  =  sec  6. 

In  the  same  way  we  find,  from  equations  (2)  of  Art.  76, 
the  system: 

[  sin  (90°  +  (9}  =  cos  0,         cos  (90°  +  ^)  =  -  sin  6, 

(2)  tan  (90°  -h  6*)  =  -  cot  6,     cot  (90°  +  ^)  =  -  tan  6, 
\  sec  (90°  +  <9)  =  -  CSC  d,     esc  (90°  +  ^)  =  +  sec  6. 


I 


172       GRAPHIC   REPRESENTATIONS   OF   FUNCTIONS 

Since  these  equations  are  true  for  angles  of  any  magnitude 
and  not  merely  for  acute  angles,  we  conclude  from  (1)  and 
(2)  that 


sin  (180°  -0)  =  sin  (90°  +  90°  -0)=  cos  (90°  -  ^)  =  sin  6, 
cos(180°-^)  =  cos(90°  +  90°-6>)=-sin(90°-^)  =  -  cos  (9, 

etc.,  giving  rise  to  the  further  system  of  equations : 

r  sin  (180°  -  ^)  =  sin  (9,  cos  (180°  -  (9)  =  -  cos  6, 

(3)        tan  (180°  -  (9)  =  -  tan  0,       cot  (180°  -  (9)  =  -  cot  (9, 
i  sec  (180°  -6)  =  -  sec  (9,       esc  (180°  -6)  =  esc  0. 

In  similar  fashion  we  find 


sin  (180°  +  ^)  =  sin(90°  +  90^  +  (9)  =  cos  (90°  +  (9)  =  -  sin  (9, 
cos  (180°  +  (9)  =  cos  (90°  +  90°  +  0)  =  -  sin  (90°  +  (9) 

=  —cos  6,  etc., 
so  that  we  obtain 


(4) 


f  sin (180°  +  (9)  =  -  sin  (9,      cos (180°  +  ^)  =  -  cos  (9, 
tan  (180°  +  <9)  =  tan  6,  cot  (180°  +  6)=  cot  0, 


sec  (180°  +  0)=-  sec  (9,       esc  (180°  +  0)  =  -  esc  ^. 

Again  we  have 

sin  (270°  -0)=  sin  (90°  +  180°  -  6)  =  cos  (180°  -  (9) 

=  —  cos  6,  etc.,    ' 
whence 

'  sin  (270°  -  ^)  =  -  cos  (9,  cos  (270°  -  (9)  =  -  sin  (9, 

(5)  tan  (270°  -0)=  cot  (9,       cot  (270°  -  <9)  =  tan  ^, 

.  sec  (270°-  0)  =  -  CSC  ^,  esc  (270°  -6)  =  -  sec  (9, 

and  similarly 

f  sin  (270°  +  <?)=-  cos  (9,  cos  (270°  +  ^)  =  sin  (9, 

(6)  tan  (270°  +  (9)  =  -  cot  (9,  cot  (270°  -^9)  =  -  tan  ^, 
[  sec  (270°  +  (9)  =  esc  ^,       esc  (270°  -\-0)  =  -  sec  (9. 

Finally  we  find  the  equations 

f  sin  (360°  -  (9)  =  -  sin  6,   cos  (360°  -  (9)  =  cos  ^, 

(7)  tan  (360°  -  (9)  =  -  tan  ^,  cot  (360°  -  <9)  =  -  cot  (9, 
I  sec  (360°  -  (9)  =  sec  l9,        esc  (360°  -  (9)  =  -  esc  d. 


THE   QUADRANTAL  FORMULA  173 

and  the  system 

sin  (360°  -\-6)=  sin  (9,    cos  (360°  +  6)=  cos  6, 

(8)  tan  (360°+  6)  =  tan  0,  cot  (360°  -{-6)=  cot  0, 
sec  (360°  + (9)=  sec  (9,    esc  (360°  +  (9)  =  esc  ^, 

which  latter  equations  merely  express  the  periodic  character 
of  the  trigonometric  functions. 

Since,  on  account  of  the  periodicity  of  the  functions,  we 
have 

sin (360°  _  ^)=  sin(-  ^  +  360°)  =  sin (-  ^),  etc., 

we  may  also  write,  in  place  of  (7), 

'  sin(— ^)=  —  sin  ^,        cos(— ^)=  cos  ^, 

(9)  tan(-  6)=  -  tan  (9,       cot(-  0)=-  cot  (9, 
.sec(— ^)=  sec  ^,  esc  (^—  0)  =  —  esc  6. 

The  48  formulae  (1)  to  (8),  the  so-called  quadrantal  formu- 
lae, have  a  very  important  practical  application.  Thei/  serve 
the  purpose  of  finding  the  values  of  the  functions  of  angles  not 
situated  in  the  first  quadrant. 

For  instance,  if  we  wish  to  find  the  sine  and  cosine  of  310^,  we  may 
use  equations  (6),  which  give 

sin  310°  =  sin  (270°  +  40°)  =  -  cos  40°, 

cos  310°  =  cos  (270°  +  40°)  =  sin  4a°. 
The  numerical  values  of  cos  40°  and  sin  40°  may,  of  course,  be  taken 
from  the  tables. 

On  account  of  the  practical  application  just  mentioned,  it 
is  important  to  be  able  to  remember  the  48  quadrantal 
formulae.  This  is  not  at  all  difficult  if  we  impress  upon  our 
minds  some  of  their  peculiarities. 

We  observe  in  the  first  place  that  all  of  the  angles  which 
appear  in  the  left  members  of  these  equations  are  of  the  form 

a  multiple  of  90°  ±  ^,  that  is,  a  cardinal  angle  ±  0, 

while  the  angle  which  appears  in  the  right  member  is  al- 
ways simply  6. 

Let  us  speak  of  those  cardinal  angles,  like  90°  and  270°, 
which  are  odd  multiples  of  90°  as  odd  cardinal  angles,  while 


174       GRAPHIC  REPRESENTATIONS  OF   FUNCTIONS 

the  even  cardinal  angles^  like  0°  and  180°,  are  even  multiples 
of  90°. 

If  now  we  look  through  our  list  of  quadrantal  formulae,  we 
observe  that  they  are  all  included  in  one  of  the  two  forms : 


(10) 


function  of  (even  cardinal  angle  ±  6) 

=  ±  same  function  of  6^ 

function  of  (odd  cardinal  angle  ±  6^ 

=  ±  corresponding  co function  of  6^ 


it  being  understood,  as  in  Art.  10,  that  the  six  functions  are 
arranged  in  three  pairs,  sine  and  cosine,  tangent  and  cotan- 
gent, secant  and  cosecant,  each  member  of  each  pair  being 
regarded  as  the  co-function  of  the  other. 

We  have  observed,  then,  that  the  same  function  occurs  in 
both  members  of  a  quadrantal  formula  whenever  the  correspond- 
ing cardinal  angle  is  even.  If  the  cardinal  angle  is  odd,  the 
function  which  appears  in  the  right  member  is  the  co  function 
of  that  one  which  appears  on  the  left. 

It  remains  to  describe  a  method  for  remembering  which  of 
the  two  signs,  +  or  — ,  should  be  used  in  any  one  of  these 
formulas.  We  may  determine  this  sign  by  thinking  of  the 
particular  case  when  ^  is  a  positive  acute  angle.  The  quad- 
rant of  the  angle  in  the  left  member  of  the  equation  will  then 
be  evident  by  inspection,  and  therefore  also  the  sign  of  the 
left  member.  The  ambiguous  sign  ±,  on  the  right  member, 
must  then  be  chosen  in  such  a  manner  as  to  make  the  two 
members  of  the  equation  agree  in  sign. 

An  example  will  make  this  clear.  We  wish  to  find  the  formula  for 
sin  (270°  +  6).     Since  270°  is  an  odd  cardinal  angle,  we  have  in  the  first 

P^^^®  sin  (270°  + (9)  =  ±  cos  ^. 

To  determine  the  sign  on  the  right  member,  we  think  of  the  case  when 
^  is  a  positiv&'acute  angle.  Then  270°+  ^  is  in  the  fourth  quadrant,  and 
sin  (270°  +  ^)  is  negative,  while  cos  0,  being  the  cosine  of  a  positive  acute 
angle,  is  positive.     Therefore  we  must  choose  the  -  sign,  so  that 

sin  (270°  +  0)  =-  cos  0, 

since  choice  of  the  +  sign  would  lead  to  the  absurdity  of  equating  a 


THE   SINE   AND   COSINE   CURVES  175 

EXERCISE   XLIII 

Express  the  following  as  functions  of  positive  acute  angles  : 

1.  cos  (-75°).  3.    tan  517°.  5.   cot  175°. 

2.  sin  325°.  4.   esc  (-412°).  6.    sec  1562°. 

7.  Express  the  functions  of  Exs.  1-6  as  functions  of  positive  acute 
angles  less  than  45°. 

8.  Compute  the  value   of  the   expression  2  sin  (3  ^  +  10°)  for  the 
values  of  ^  =  25°,  50°,  75°,  100°. 

Find  the  values  of  the  following  expressions  : 

9.  cos  60°  cos  120°  -  sin  60°  sin  120°. 

10.  sin  30°  cos  300°  +  cos  30°  sin  300°. 

11.  tan  11^  tan  li^  +  cot  1^1^^]  cot  (^^ . 

6  3  \      6      y        V    3     / 

12.  cos  315°  sin  11°  -  tan  293°  sec  25°. 
Simplify  the  following  expressions  : 

13.  a2  +  &2  +  2  a&  cos  (180°  -  x). 

14.  (a  -  b)  tan  (90°  -{-  A)  +  (a  +  b)  cot  (-  A). 

15.  asml^-0]  +  bcos(7r-0). 

16.  tan  6  +  tan  (tt  -  6). 

State  for  what  values  of  0  each  of  the  following  expressions  is  positive, 
and  for  what  values  of  ^  it  is  negative : 

17.  sin  ^- cos  a  19.    sin2  ^  -  cos^  ^. 

18.  sin  0  +  cos  0.  20.    tan  0  -  cot  6. 

21.  Find  the  formulae  for  the  functions  of  ^  —  ^  in  terms  of  the 
functions  of  0,  the  angle  6  being  measured  in  radians. 

22.  Find  formulae  for  the  functions  of  ^-tt  in  terms  of  the  functions 
of  ^. 

78.  Properties  of  the  sine  and  cosine  curves.  The  prop- 
erties of  the  sine  and  cosine  which  were  discussed  in  Arts. 
75,  76,  77  manifest  themselves  very  clearly  if  we  make  use 
of  the  graphs  of  these  functions  as  obtained  in  Art.  78.  We 
shall  slightly  modify  these  graphs,  however,  by  thinking  of 


176       GRAPHIC   REPRESENTATIONS  OF  FUNCTIONS 


the  angle  x  as  being  measured  in  radians  (see  Art.  74)  rather 
than  in  degrees,  and  by  choosing  the  same  unit  of  length  to 


+?/ 


> 

J- 

P" 

-/ 

C' 

V' 

4' jjf' 

°/ 

^ 

A 

Px 

B      M"\C 

-A- 

l^TT 

-^ 

X 

~  2 

y 

0 

M 

EM" 
2 

3fr 

7^ 

'         F' 

p> 

+x 


Fig.  100.  —  The  Sine  Curve.    Natural  Scale 

represent  one  radian  on  the  a^-axis  as  that  which  represents 
the  abstract  number  1  on  the  i/-axis. 

Figure  100  represents  the  sine  curve 
y  =  sin  a: 
constructed  in  accordance  with  this  choice  of  units. 

Let  us  consider  two  points  of  this  curve  which  are  at  the 
same  distance  from  the  ^-axis  but  on  opposite  sides,  such  as 
P  and  P^  The  ordinates  of  these  points  are  obviously 
numerically  equal  but  opposite  in  sign.  Denote  the  abscissa 
of  P  by  2 ;  then,  that  of  P'  will  be  —  2,  and  we  find  that 
(1),  sin  (  —  2)  =  —  sin  z. 

If  we  draw  a  line  parallel  to  the  y-axis  through  the  point 

A  for  which  a;  =  — ,  this  line  clearly  divides  the  curve  into 

two  symmetrical  portions.     Consequently  two  points,  such 
as  P  and  P'^  at  the  same  distance  from  this  line  but  on 
opposite  sides  of  it,  will  have  equal  ordinates.     That  is, 
sin  Oif  =  sin  OM^K 
But  if  we  denote  MA  by  z,  we  have 

OM^'^-z,     OM" 
so  that 
(2).  sin(|-2)  =  sin(|  +  ^). 


TT 


+  z, 


THE   SINE   AND  COSINE   CURVES  177 

Let  the  curve  be  divided  into  two  portions  by  a  line  parallel 
to  Oy  through  the  point  B  whose  abscissa  is  equal  to  tt. 
Points  of  the  curve  at  equal  distances  from  this  line  but  on 
opposite  sides  of  it,  such  as  P^'  and  P"\  have  ordinates 
numerically  equal  but  opposite  in  sign.     Therefore 

(3),  sin(7r— 2)  =  —  sin(7r -f  2). 

In  similar  fashion  we  find 

(4).  sin  {-^  -  2 j  =  sin  (^  +  ^j, 

(5),  sin(2  7r— 0)  =  — sin(2  7r  +  2). 

The  points  M  and  M"  were  equidistant  from  A.  Conse- 
quently the  distances  OM  and  M'^B  are  equal.  Since  the 
ordinates  MP  and  M"P"  are  equal,  we  shall  have 

sin  OM"  =  sin  OM. 

If  we  put  0M=  2,  we  shall  have  M" B  =  z^  and  therefore 

OM''  =  TT  -  2, 

so  that  the  preceding  equation  becomes 

(6),  sin  Qir  —  z)  =  sin  z. 

By  similar  considerations  in  connection  with  the  cosine 
curve,  we  find  a  system  of  relations  which  correspond  com- 
pletely to  the  above  equations  (1),  to  (6),.     They  are 

(l)c  cos  (  —  2)  =  cos  2, 

(2)c  cos(|-^)  =  -cos(|  +  2), 

(3)c  COS  (tt  —  2)  =  cos  (tt  +  ^), 

(4\  cosf-^-2J  =  -  cosf-^H-2J, 

(5)c  cos(2  7r  — 2)  =  cos(2  7rH- 2), 

(6)c  cos  (tt  —  2)  =  —  cos  2. 

The  truth  of  all  these  equations  which  are  thus  suggested 
by  the  curves  has  been  established,  in  a  slightly  different 
notation,  in  Arts.  75-77. 


178      GRAPHIC   REPRESENTATIONS  OF   FUNCTIONS 

We  can  hardly  fail  to  notice  the  striking  similarity  between 
the  sine  and  cosine  curves.  In  order  to  put  into  evidence 
the  relations  between  them,  we  construct  Fig.  101  which 
contains  them  both. 

+ 


+  07 


Fig.  101 


This  figure  suggests  that,  if  we  displace  the  cosine  curve 


TT 


toward  the  right  through  a  distance  of  —  units,  it  will  coin- 

cide  with  the  sine  curve.     If  this  is  true,  a  point  P  of  the 
cosine  curve  whose  coordinates  are 


(a) 


0M=  2,    il!fP  =  C0S2J, 


and  which  by  our  displacement  will  be  brought  into  coin- 
cidence with  a  point  P^  whose  coordinates  are 


iV) 


OM'  =  z  +  ^,  M'P'  =  MP, 


should,  in  its  new  position,  be  a  point  on  the  sine  curve. 
Therefore  the  coordinates,  OW  and  M^P\  of  this  point  P^ 
should  satisfy  the  relation  y  —  sin  x  which  is  satisfied  by  the 
coordinates  of  all  points  of  the  sine  curve.     This  gives 


(O  MP^  =  sin  [z  +  1^, 

and,  on  combination  with  (a)  and  (5), 

(7),  sin  ^2  +  I")  =  M^P^  =  MP=  cos  z. 


THE   SINE   AND   COSINE   CURVES  179 

Thus,  equation  (7),  must  be  true  if  the  geometric  relation 
between  the  sine  and  cosine  curves  suggested  by  Fig.  101  is 
actually  based  on  fact.  But  this  equation  coincides  with 
formula  (2)  of  Art.  76,  except  for  the  notation,  so  that  its 
validity  is  no  longer  open  to  question. 

Consequently,  the  sine  and  cosine  curve  differ  only  in  posi- 
tion and  may  he  brought  into  coincidence  by  a  displacement  of 

—  units  parallel  to  the  x-axis. 


EXERCISE  XLIV 

1.  What  geometric  property  of  Fig.  101  corresponds  to  the  relations 

sin  f  -  —  2  ]  =  cos  z,  cos  (^  —  2=)  =sin2? 

2.  How  are  the  relations  ,  -,  ^ 

sin  f  —  +  2  j  =  -  cos  2,  cos  (  -^  +  2;  J  =  sin  2 

to  be  obtained  from  Fig.  101  ? 

3.  Plot  the  tangent  and  cotangent  curves  and  discuss  these  graphs  in 
a  fashion  analogous  (so  far  as  possible)  to  the  discussion  of  Art.  78. 


CHAPTER  XI 


RELATIONS   BETWEEN   THE  FUNCTIONS  OP  MORE   THAN 

ONE  ANGLE 

79.   The  addition  theorems  for  sine  and  cosine.     In  Art.  77 

we  expressed  sin  (  ^  +  ^  J,  cos  ( ^  +  ^  )i  etc.,  in  terms  of  sin  6 

and  cos  0,     The  angle  0  was  an  angle  of  any  magnitude,  but 

the  angle  added  to  it  was  always  an  integral  multiple  of  — 

radians  or  90°.  The  question  now  arises  whether  it  is  pos- 
sible to  find  similar  formulse  for  sin  (a  +  /3)  and  cos  (a  H-  ^), 
where  both  a  and  /3  are  angles  of  any  magnitude. 

Let  us  assume,  to  begin  with,  that  a  and  ^  are  both  positive 
acute  angles  whose  sum  a  +  yS  is  also  acute.  We  place  the 
acute  angle  a  in  its  standard  position 
xOA.  (See  Fig.  102.)  We  then 
place  the  angle  ^  with  its  initial  side 
upon  OA  (the  terminal  side  of  the 
angle  a),  so  as  to  make  Z  A  OB  equal 
to  yS.     Then 

ZxOB=  «  + A 
and  moreover  this  angle  is  in  its  stand- 
ard position.     Therefore  if   we    take 
any  point  P,  different  from  0,  on  its  terminal  side  OB  and 
drop  a  perpendicular  PM  from  P  to  the  a;-axis,  we  shall  have 

(1)       '       sin  («  +  yS)  =  ^,  cos  («  +  ^)  =M; 

Let  us  drop  perpendiculars  FQ,  §iV,  and  QE  from  P  to 
OA,  from  Q  to  the  a;-axis,  and  from  Q  to  MP.     Then  we  have 
MP 


^x 


FiQ.  102 


\ 


(2)       sin  («  +  yS)  = 


OP 


NQ-\-RP  ^  NQ      RP 
OP  OP      OP' 


180 


ADDITION  THEOREMS  FOR  SINE  AND   COSINE      181 

NO 

Now  — ^  is  a  ratio  of  sides  of  two  different  right  triangles^ 

namely,  ONQ  and  OPQ.     But  these  triangles  have  the  side 
OQ  m.  common,  and  this  common  side  may  be  used  to  trans- 

form  — ^  into  a  product  of  two  ratios,  each  of  which  contains 

two  sides  of  the  same  right  triangle  and  is  therefore  a  trigo- 
nometric function  of  the  acute  angles  of  this  triangle.     In 
fact  we  find 
.o.  NQ     NQ     OQ        .  Q 

In  the  same  way  we  find  for  the  second  term  of  (2) 

RP^RP    FQ 
OF      PQ  '  OP' 


But 


PO  HP 

—^  =  sin  /9,  and  — —  =  cos  BPQ  =  cos  a, 


since  the  sides  of  the  angle  BPQ  are  respectively  perpen- 
dicular to  those  of  a.     Consequently  we  find 

(4)  -^  =  cos  a  sin  fi. ) 

If  (3)  and  (4)  be  substituted  in  (2),  we  obtain  the  impor- 
tant formula 

(5)  sin  (a  +  yS)  =  sin  a  cos  yS  +  cos  a  sin  yS. 
Referring  once  more  to  Fig.  102,  we  have 

(6)  cos(.4-^)=^=^^--^g  =  ^-^. 
^  ^  ^        ^^       OP  OP  OP      OP 

If  we  again  transform  each  of  these  ratios  into  a  product 
of  two  others,  we  find 

ON      ON     OQ  r, 

RQ      RQ    PQ        .         .     r, 


182 


FUNCTIONS  OF  MORE   THAN  ONE  ANGLE 


so  that 

(7)  cos  (a  4-  ;Q)  =  cos  a  cos  yS  —  sin  a  sin  yS. 

It  is  easy  to  see  that  equations  (5)  and  (7)  remain  valid  if 
a  and  yS  are  acute  angles,  even  if  their  sum  is  greater  than  a 

right  angle.  In  that  case  we  shall 
have  the  situation  represented  in 
Fig.  103.  If  we  make  precisely  the 
same  constructions  as  before,  the 
proof  of  the  formula  for  sin  (a  +  yS) 
will  remain  applicable  word  for  word. 
But  since  a  +  /3  is  now  in  the  sec- 
ond quadrant,  its  cosine  is  negative  ; 
that  is, 

,        ^r,^  MO 

COS  (a  +  /5)  =  -  — , 

where  by  MO  we  mean  merely  the  positive  number  express- 
ing the  length  of  the  line-segment  MO^  so  that  —MO  is  a 
negative  number. 

Now  the  figure  shows  that 

MO  =  MN-  0N=  RQ-  ON, 
so  that 

MO^ON-RQ^ON     RQ 

OP      OP' 


cos  (a  -I-  y3)  = 


OP  OP 

From  this  point  on,  the  proof  proceeds  exactly  as  in  the  pre- 
vious case,  beginning  from  equation  (6). 

Thus,  we  have  proved  that  equations  (5)  and  (7)  are  cer- 
tainly true  if  a  and  y8  are  positive  acute  angles,  even  if  their 
sum  is  greater  than  90°. 

We  may  now  show  that  these  formulae  are  true  for  two 
angles  in  any  quadrant.  In  o«^er  to  do  tir  ,.  we  first  prove 
the  following  theorem.     If  the  formulcB 

(8)  sin  (a  -f-  y3)  =  sin  a  cos  yS  4-  cos  a  sin  ^, 

cos  (a  -f-  yS)  =  cos  a  cos  /8  —  sin  a  sin  y9, 

are  true  for  two  angles  a  and  yS,  they  remain  true  if  either  of 
these  angles  he  increased  hy  90°. 


ADDITION"  THEOREMS  FOR  SINE   AND   COSINE      183 

Proof.  Let  us  assume  that  equations  (8)  are  true  for  a 
certain  pair  of  angles  a  and  /3.     Put 

a' =  90°+ a, 
so  that  . 

«'  +  ^  =(90°  +  «)+  ^. 

We  shall  then  have  (Art.  77,  equations  (2)), 

sin  {a!  +  /3)  =  sin  (90°  +  a  +  yS)  =  cos  (a  +  y8) 
(9)  =  cos  a  cos  /3  —  sin  a  sin  yS, 

cos  (ce^  +  /3)  =  cos  (90°  +  a  +  /8)  =  -  sin  (a  +  yQ) 

=  —  sin  a  cos  yS  —  cos  a  sin  )S, 
and  also 

sin  a!  =  sin  (90°  +  a)  =  cos  a,  cos  a!  =  cos  (90°  +  a)  =  —  sin  a, 

whence 

sin  a  =  — cos  a',        cos  a=  sin  a'. 

If  we  substitute  these  values  in  (9),  we  find 

sin  (a'  -{-  y8)  =  sin  a'  cos  yS  +  cos  a'  sin  yS, 
cos  (a'  -}-  y8)  =  cos  a'  cos  yS  —  sin  a'  sin  yS. 

But  these  formulae  are  of  the  same  form  as  equations  (8), 
with  a'  =  a  +  90°  in  place  of  a.  The  same  process  would 
show  that  equations  (8)  would  still  be  satisfied  if  we  replaced 
y(3  by  y8^  =  90°  +  y8.     Consequently  our  theorem  is  proved. 

We  know  already  that  equations  (8)  are  true  if  a  and  yS 
are  any  two  positive  acute  angles.  On  account  of  the 
theorem  just  proved  they  will  still  be  true  if  a  is  in  the 
second  quadrant  and  y8  in  the  first,  and  hence  also  if  both  a 
and  y8  are  in  the  second  quadrant,  and  hence  if  one  of  these 
angles  is  in  the  third  quadrant,  etc.  Therefore  equations 
(8)  are  true  fo^    dl  positive  angles. 

But  they  are  also  true  if  either  or  both  angles  are  negative. 
For  instance,  let  a  be  a  negative  angle  while  yS  is  positive. 
By  adding  to  a  a  sufficient  number  n  of  complete  positive 
revolutions,  we  shall  obtain 


I 


a'  =  a +  71.  360°, 


184    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 

a  positive  angle.     But  then 

sin  a'  =  sin  a,  cos  a'  =  cos  a, 

sin  (a'  +  y8)  =  sin  (a  +  yS),  cos  (a'  -h  y8)  =  cos  (a  +  yS), 

so  that 

sin  (a  +  y8)  =  sin  (a'  +  /3)  =  sin  a'  cos  /8  +  cos  a'  sin  /3 

=  sin  a  cos  yS  -j-  cos  a  sin  /3, 

and  similarly  for  cos  (a  +  z^)- 

If  yS  is  negative,  we  may  proceed  in  the  same  manner. 
Consequently  we  obtain  the  following  important  theorem: 

If  a  and  ^  are  two  positive  or  negative  angles  of  any  magni- 
tude^ the  sine  and  cosine  of  their  sum  are  always  given  hy  the 
formulcB 

sin  (a  +  P)  =  sin  a  cos  p  +  cos  a  sin  p, 

(8) 

COS  (a  +  p)  =  cos  a  cos  p  —  sin  a  sin  p. 

The  method  of  mathematical  induction  employed  for  proving 
the  general  validity  of  these  formulae  is  of  fundamental 
importance  in  all  parts  of  mathematics.  We  have  already 
made  use  of  it  in  Art.  76. 

Equations  (8)  are  usually  known  as  the  addition  theorems 
for  sine  and  cosine. 

EXERCISE   XLV 

1.  Cornpute  the  sine  and  cosine  of  75*^. 

Solution.  We  have  75°  =  30°  +  45°,  sin  30°  =  ^  cos  30°  =  ^  V3,  sin  45'=' 
=  cos  45°  = Therefore 

sin  75°  =  sin  30°  cos  45°  +  cos  30°  sin  45°  =  ? 
cos  75°  =  cos  30°  cos  45'  -  sin  30°  sin  45°  =  ? 

Compute  the  sines  and  cosines  of  the  following  angles : 

2.  120°.  4.   210°.  6.   195°. 

3.  150°.  5.   105°.  ~7.   165°. 
^8.   Find  formulae  for  sin  (45°  +  6)  and  cos  (45°  +  6). 

\J  9.   Find  formulae  for  sin  (30°  +  &)  and  cos  (30°  -\- &). 


THEOREMS  FOR  TANGENT  AND  COTANGENT   185 

x^O.    Find  formulae  for  sin  (60°  +  0)  and  cos  (60°  +  0). 

11.  Show  that  those  of  the  quadrantal  formulae  of  Art.  77,  which  in- 
volve the  sine  and  cosine,  are  special  cases  of  the  addition  theorems  for 
the  sine  and  cosine. 

Prove  that  the  following  equations  are  true  for  all  values  of  the  angles 
which  appear  in  them.     That  is,  prove  that  they  are  identities. 

^12.   sin  (a  +  (3)  cos  /3  +  cos  (a  +  /3)  sin  ^  =  sin  {a +  2  (3). 

^13.   cos  (a  +  y8)  cos  ^  —  sin  (a  +  /3)  sin  yS  =  cos  (a  +  2  /?). 

f  14.  Show  how  sin  (a  +  /8  +  y)  and  cos  («  +  /3  +  y)  may  be  expressed 
in  terms  of  the  functions  of  a,  (3,  and  y. 

V  Simplify  the  following  expressions  : 

15.    sin  (1  +  w)  ^  cos  (1  -  n)  6  +  cos  (1  +  n)  ^  sin  (1  -  n)  0. 
'   16.    cos  (1  -f  n)  ^  cos  (1  -n)  ^  -  sin  (1  -{-  n)  6  sin  (1  -  n)  6. 

80.  The  addition  theorems  for  tangent  and  cotangent.     The 

addition  formulae  for  the  tangent  and  cotangent  may  be 
obtained  as  consequences  of  those  for  the  sine  and  cosine. 
We  have 

tan  C    4-  3^=:  sii^  («  +  )^)  _  sin  a  cos  /3  +  cos  ct  sin  /3  ^ 
cos  (a  +  /3)      cos  a  cos  /5  —  sin  a  sin  /3 

The  fraction  in  the  right  member  of  this  equation  may  be 

expressed  in  terms  of  tan  a  and   tan  yQ  by  dividing  both 

numerator  and  denominator  by  cos  a  cos  yS.     We  obtain  in 

this  way 

sin  a  cos  ff      cos  a  sin  /S 

^         ON       COS  a  cos  S      cos  a  cos  8 

tan  (a  +  yS)  = ^ 3 ^, 

-J  _  sin  a  sin  yo 

cos  a  cos  yS 

or,  finally, 

(1)  tan(a  +  B)=   tang  +  tanp^ 

^         '^^      1-tana  tan  p 

By  a  similar  process  we  find 

r9\  ^^4-  r     I    /D\      cot  a  cot  yS  —  1 

(2)  cot(a  +  ^)  = ^         . 

cot  a  +  cot  yO 


186    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 


EXERCISE   XLVI 

Find  the  tangents  and  cotangents  of  the  following  angles : 
'^  1.   75°.  3.  210°. 

2.  105^  4.    150°. 

Prove  that  the  fdllowing  equations  are  identities: 
1  +  tan  ^ 


5.  tan  (45°  +  6) 

6.  cot  (45°  +  6)  ■■ 


1  -  tan  d 
cot  ^  -  1 


cot  ^  +  1 

7.  Find  formulae  for  tan  (30°  +  0)  and  cot  (30°  +  6). 

8.  Find  formulse  for  tan  (60°  +  0)  and  cot  (60°  +  0). 

81.  The  subtraction  formulae.  The  formulse  for  sin  (a  —  /S), 
cos  (a  —  ^),  etc.,  may  be  obtained  from  those  for  sin  (a  +  13)^ 
cos  (a  +  yS),  etc.  For,  since  the  addition  formulse  are  known 
to  be  true  for  all  positive  and  negative  values  of  a  and  jS,  we 
may  write  ^^  _  ,    ,m;S  -^^-^  ^>^^i'i^ 

sin  (a  -  /3)  =  sin  [a  +  (-  yS)] 

=  sin  a  cos  (—  yS)  +  cos  a  sin  (—  yQ), 

cos  (a  —  fi)  =  cos  [a  +  (  —  ^)] 

=  cos  «  cos  (—  /3)  —  sin  a  sin  (  —  yS). 

But,  on  account  of  equations  (9),  Art.  77,  we  have 

sin  (  —  /3)  =  —  sin  A    cos  (  —  yS)  =  cos  ^, 

so  that  we  find 

sin  (a  —  p)  =  sin  a  cos  p  —  cos  a  sin  p, 

(1) 

cos  (a  —  p)  =  cos  a  cos  p  -f  sin  a  sin  p. 

In  the  same  way  we  find 

^ox  ^       ^        Q\       tana  — tan  p 

(2^  tan  ( a  —  B)  = ^t 

^  ^  V        r/      i+tanatanp 

and 

xox  4.  ^        o\      cot  a  cot  yS  +  1 

(3)  cot(a— p)= ^-- . 

^^  ^    .   "^^        cot  y8- cot  a 

Of  course  equations  (2)  and  (3)  may  also  be  obtained 
from  (1)  by  division. 


PRODUCTS  AND   SUMS  OF  FUNCTIONS  187 

EXERCISE    XLVII 

1.  Let  the  student  devise  a  geometric  proof  for  equations  (1),  in  the 
case  when  a  and  y8  are  positive  acute  angles,  a  being  the  larger  of  the  two. 

Note.  If  Fig.  102  be  described  in  words,  but  if  the  angle  there 
denoted  by  fi  be  instead  regarded  and  constructed  as  a  negative  angle, 
the  same  description  will  apply  to  both  figures  and  the  same  method  of 
transforming  the  ratios  which  was  used  in  Art.  79  will  be  effective  in  the 
present  example. 

82.  Formulae  for  converting  products  of  trigonometric  fimc- 
tions  into  siuns,  and  vice  versa.  Before  the  invention  of 
logarithms  the  calculations  of  products  and  quotients  was  a 
very  laborious  process.  In  those  days,  then,  it  was  considered 
a  great  simplification  if  a  formula  whose  numerical  evaluation 
required  multiplication  could  be  transformed  into  another 
requiring  only  addition  or  subtraction.*  The  addition  and 
subtraction  formulfe  will  enable  us  to  accomplish  this  for  a 
product  of  two  sines,  of  two  cosines,  or  of  a  sine  and  cosine. 

In  fact,  we  have  the  two  equations 

sin  (a  -j-  )S)  =  sin  a  cos  ^  -h  cos  a  sin  ^, 
""  sin  (a  —  yS)  =  sin  a  cos  y8  —  cos  a  sin  yS, 

from  which  we  derive,  by  addition  and  subtraction, 

^1 X  sin  (a  +  /3)  -|r  sin  (a  —  /3)  =  2  sin  a  cos  y8, 

sin  (a  +  /3)  —  sin  (a  —  /S)  =  2  cos  a  sin  yS. 

In  the  same  way,  from 

cos  (a  4-  /3)  =  60S  a  cos  yS  —  sin  a  sin  /8, 
cos  (a  —  /3)  =  cos  a  cos  y8  +  sin  a  sin  /3, 

we  find 

>^2\  cos  (a  +  /3)  +  cos  (a  —  /S)  =  2  cos  a  cos  yS, 

cos  (a  +  /3)  —  cos  (a  —  yS)  =  —  2  sin  a  sin  yS. 

From  these  last  equations  we  find,  by  trUnsposition  and 
division  by  2, 

^gv  cos  a  cos  y8  =  1  [cos  (a  —  /3)  +  cos  (a  +  y8)], 

sin  a  sin  /8  =  J  [cos  (a  —  y8)  —  cos  (a  +  /3)], 

*  This  was  the  so-called  prosthaphxretic  method. 


188    FUNCTIONS  OF  MORE  THAN  ONE  ANGL^ 

whereas,  from  the  first  equation  of  (1),  we  find 

(4)  sin  a  cos  yS  =  ^  [sin  («  —  yS)  +  sin  (a  +  y8)] . 

The  result  obtained  from  the  second  equation  of  (1)  is 
the  same  as  (4)  except  for  an  interchange  of  the  letters  a 
and  /3. 

liquations  (3)  and  (4)  are  important  in  that  they  enable 
U8  to  transform  a  product  of  two  sines^  two  cosines^  or  of  a  sine 
and  a  cosine  into  a  sum  or  difference. 

For  many  purposes  this  is  very  important  even  nowadays,* 
although  not  for  the  purposes  of  numerical  calculation.  In 
fact,  at  the  present  time,  for  numerical  work  we  prefer 
formulae  which  involve  multiplication  to  those  involving 
addition,  because  the  former  process  is  more  easily  performed 
by  logarithms. 

The  above  formulae,  slightly  modified,  may  also  be  used 
for  the  purpose  of  converting  sums  and  differences  of  sines 
and  cosines  into-  products.     Let  us  put 

a  +  yS  =  ^,  a-ff=B, 

sothat  a  =  i(^  +  ^),  /3  =  l(^-5). 

Then,  equations  (1)  and  (2)  yield  the  following  four  for- 

mulie  : 

'sin  ^  +  sin  ^  =  2  sin  i-  (A  +  B)  cos  J  (A  -  J?), 
sin  ^  -  sin  ^  =  2  sin  ^{A  —  B}  cos  i(A-\-  j5), 
cos  ^  +  cos  ^  =  2  cos  -|-  (A  —  B)  cos  -J  (A  +  B), 
cosA-GosB  =  -2  sin^^A  -  B)  sin  -|-  (J.  +  B). 

We  have  seen  the  first  two  of  these  equations  before,  hav- 
ing derived  them  directly  from  a  figure  (in  Art.  49)  for  the 
purpose  of  proving  the  law  of  tangents.  Our  proof,  at  that 
time,  did  not  permit  us  to  affirm  that  the  formulae  were  true 
for  all  angles  A  and  B.  That  such  is  the  case,  however,  has 
now  been  made  evident,  since  the  present  proof  was  obtained 
without  placing  any  restrictions  on  the  values  of  the  angles 
A  and  B. 

*  For  instance,  in  harmonic  analysis  (see  Arts.  Ill  and  112)  and  in  the  integral 
calculus. 


(5) 


FUNCTIONS  OF   DOUBLE  ANGLES  189 

EXERCISE  XLVIII 

Prove  the  following  equations: 

1.  sin  3  ^  +  sin  ^  =  2  sin  2  6  cos  6. 

2.  sin  (^+x]  +  sml--x\  =  2  sin  -  cos  a:  =  V2  cos  x. 

«     sin  6  a  -f  sin  4  a      ,       e 

3.    ^ =  tan  5  a. 

cos  6  a  +  cos  4  a 

.     sin  a  —  sin  /3 2 

sin  a  +  sin  yS  ~  ^^^  a  +  ^' 
2 

^  5.  cos  a  —  cos  3  a  =  2  sin  a  sin  2  a. 

6.  sin  3  a  +  cos  a  =  sin  3  ct  +  sin  (90°  -  a)  =  ? 

7.  By  generalizing  the  process  observed  in  Ex.  6,  derive  a  formula 
for  sin  a  +  cos  ft. 

/^.  Derive  a  formula  for  sin  a  —  cos  ft. 

Reduce  the  following  products  to  sums  or  differences : 
9.   sin  4  a  cos  2  a.  12.   cos  2  ^  cos  8  6. 

10.  sin  6  ^  sin  4  9.  13.   sin  5  a  cos  3  a. 

11.  cos  2/3  sin  4^.  -i-14.   sin^^cos^. 

15.   Making  use  of  the  formulae  (5)  of  Art.  82,  show  how  to  derive 
the  law  of  tangents  from  the  law  of  sines. 

83.   Functions  of  double  angles.     If  we  put  yS=  a  in  equa-N 

tions  (8)  of  Art.  79,  we  find 

(1)  sin  2  a  =  2  sin  a  cos  a, 

and 

(2)  cos  2  a  =  cos^  a  —  sin*'*  a.     ' 

On  account  of  .the  relation 

sin^  a  +  cos^  a  =  1, 

the  latter  equation  may  also  be  written  in  either  of  the  fol- 
lowing two  forms : 

(3)  cos  2  a  =  1  —  2  sin^  «, 
or 

(4)  cos2a=  2cos2a— 1. 


190    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 

If  we  put  fi  =  ain  equations  (1)  and  (2)  of  Art.  80,  we 
find 

2  tan  a 


(5)  tan  2  a  = 

(6)  cot  2  a  = 


1  —  tan^  a 
cot^  a  —  1 


2  cot  a 

which  equations  may,  of  course^also  be  derived  from  (1)  and 
(2)  by  division. 

84.  Functions  of  half  angles.  In  Art.  83  we  regard  the 
functions  of  a  as  known,  and  we  learn  how  to  compute  the 
functions  of  2  a.  We  shall  now  invert  the  problem  by 
regarding  as  known  the  functions  of  2  a,  the  problem  being 
to  calculate  the  functions  of  a,  the  half  angle.  To  put  the 
character  of  the  problem  more  clearly  into  evidence,  we  shall 

which  merely  amounts  to  thinking  of  any  angle  ^  as  a  double 
angle;  namely,  as  double  its  half. 

With  this  change  of  notation,  equations  (3)  and  (4)  of 

Art.  83  become 

(1)  cos  6>  =  1  -  2  sin2  |,         cos  (9  =  2  cos2  ^  -  1. 

e 

If  we  solve  the  first  of  these  equations  for  2  sin^  q  ^^^  ^^® 
second  for  2  cos^  -,  we  find 

(2)  2sin2^  =  l-cos(9,        2  cos2^=  1  +  cos  (9, 
whence 

ro\      z   Q     .  ^  /i  -  cos  e  e     ,  ^/i  +cose 

(3)  sin  — =±\/ ,        cos-=±\/^ 

^^  2^2  2^2 

The  ambiguous  signs  on  the  right  members  are  determined 

9 
by  the  quadrant  of  the  angle  -•    If  ^  is  a  positive  angle  not 

6 
greater  than  180°,  -  is  in  the  first  quadrant  and  the  +  sign 

must  be  chosen  in  both  of  the  equations  (3). 


FUNCTIONS  OF   HALF  ANGLES  191 

From  (3)  we  find,  by  division, 


C4)        tan  1  =  ±  Ji^^^,       cot  ^  =  ±  V^^t^^, 
^  2  Vl  +  cosG  2  ^1-cose 

the  appropriate  sign  being  again  determined  by  the  quadrant 
of  the  angle  J  0, 

According  to  (1),  Art.  83,  we  have 

6        6 
(5)       *  2  sin  -  cos  -  =  sin  6. 

Let  us  divide  each  member  of  the  first  equation  of  (2)  by 
the  corresponding  member  of  (5).     We  find 

^^.x  ,      0     1  —  cos  0 

(P)  tan-  =  — r-^ — 

2         sm  d 

By  a  similar  process  we  obtain,  from  the  second  equation 
of  (2), 

(7)  cot|  =  l±^. 

2         sm  ^ 

These  last  two  formulae  might  also  have  been  derived  from 
(4).  But  the  proof  we  have  given  is  preferable  since  it 
avoids  the  necessity  of  discussing  the  ambiguous  sign,  a  dis- 
cussion which  would  be  necessary  if  we  had  followed  the 
other  method. 

The  equations  for  sin  -,  cos  -,  etc.,  are  of  very  great  impor- 

tance,  because  theymay  he  used  for  the  purpose  of  computing  a 
table  of  trigonometric  functions.  In  fact,  we  have  already 
shown  how  to  calculate  the  values  of  the  functions  of  0°,  30°, 
45°,  60°,  and  90°  (Art.  11).  By  means  of  the  addition  and 
subtraction  formulae  (Arts.  79,  81)  we  are  therefore  in  a  posi- 
tion to  find  also  the  functions  of  15°  and  75°.  We  can  now 
calculate  the  values  of  the  functions  of  one  half  of  15°  or 
7°.5,  of  one  half  of  7°.  5,  or  3°.75,  etc.  By  continuing  this 
process  of  bisection  and  combining  the  results  by  means  of 
the  addition  theorem,  we  may  obviously  compute  the  values 
of  the  functions  for  a  set  of  angles  between  0°  and  90°  as 


,%^ 


192    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 

close  together  as  we  please.     By  interpolation  we  may  then 
find  the  functions  of  1°,  2°,  3°,  etc. 

The  method,  of  which  we  have  just  given  an  outline,  is  essentially  the 
same  as  that  employed  by  Ptolemy  (second  century  a.d.).*  Ptolemy, 
however,  also  made  use  of  the  inscribed  pentagon  (cf.  Exercise  XLIX, 
Ex.  12),  and  his  table  was  a  table  of  chords,  not  of  sines.  (See  Art.  70.) 
His  table  gives  the  values  of  the  chord  for  each  half  degree  of  arc  with  a 
degree  of  accuracy  somewhat  greater  than  that  which  would  correspond 
to  a  modern  five-place  table.  The  earlier  tables  of  Hipparchus  and 
Menelaus  are  not  extant. 

The  Hindus  followed  the  method  which  we  have  outlined  even  more 
closely.  In  fact,  the  table  given  by  Aryabhata  (born  476  a.d.)  gives 
the  values  of  the  sine  at  intervals  of  3°  45'.  As  we  have  seen,  this  is 
precisely  the  interval  which  would  arise  as  a  result  of  continued  bisec- 
tion of  30°. 

Essentially  the  same  method  was  used  in  subsequent  improvements 
and  enlargements  of  these  tables,  especially  by  Rheticus  (1514-1574) 
and  PiTiscus  (1561-1613).  Other  far  more  powerful  methods  have 
since  been  developed,  based  essentially  on  the  notions  of  the  calculus 
and  the  theory  of  infinite  series. 

EXERCISE    XLIX 

1.  From  the  functions  of  30°  find  those  of  60°. 

2.  From  the  functions  of  60°  find  those  of  120°. 

3.  From  the  functions  of  90°  find  those  of  45°. 
•  4.   From  the  functions  of  30°  find  those  of  15°. 

5.   From  the  functions  of  15°  find  those  of  7°.5. 
i   6.   Find  formulae  for  sin  3  a,  cos  3  a,  tan  3  a. 
"^HiNT.     Put  3  a  =  2  a  +  oT^ 

7.  Find  formulae  for  sin  4  a,  cos  4  a,  tan  4  a. 

8.  Find  formulae  for  sin  5  a,  cos  5  a,  tan  5  a. 

9.  Prove  formula  (1),  Art.  83,  by  means  of  a  figure. 

I    ^10.    Given  tan  $  =  ^,  0  being  in  the  first  quadrant.     Find  the  func- 
tions of  2  ^  and  ^  0. 

*  Ptolemy's  great  work  on  Astronomy,  usually  known  as  the  Almagest,  re- 
mained in  undisputed  authority  until  the  time  of  Copernicus.  The  so-called  Ptole- 
maic system  of  astronomy,  as  opposed  to  the  more  modern  Coperuicau  system 
was  named  after  him  for  this  reason. 


I 


CERTAIN  LIMIT   RELATIONS  193 

''   11.    If  0  is  in  the  third  quadrant  and  sin  0  =  — —,  find  the  functions 

Df  2  e.  ^5 

12.  In  a  circle  of  radius  1,  inscribe  a  regular  pentagon.  Show  that, 
by  means  of  this  construction  the  trigonometric  functions  of  72"^  and  18° 
may  be  computed.     In  particular,  show  that 

8inl8°  =  |:(V5-l). 

13.  Making  use  of  the  results  of  Ex.  12,  compute  the  functions  of  12°. 

14.  Making  use  of  the  results  of  Ex.  13,  compute  the  functions 
of  6°. 

Assuming  the  truth  of  the  law  of  cosines,  and  setting  s  =  l{a  +  1)  -^  c), 
prove  the  following  formulae  for  the  functions  of  the  half  angles  of  a 
triangle. 


15.   sinM=A/^'^~^^^'~"^ 
^  be 


16.  cosi^=V'^^^^^ 
^      he 


17.  taDM=\K-'-''>^-''-">- 

^      s(s  -  a) 

18.  Prove  formulae  (6)  and  (7)  of  Art.  84  by  means  of  equations  (4). 

85  a.  The  limit  ^^-  and  related  limits.  Although  the  method 

sketched  in  Art.  84  for  calculating  a  table  of  the  values  of 
the  trigonometric  functions  is  adequate,  it  involves  far  more 
labor  than  is  actually  necessary.  The  following  theorem, 
whose  truth  is  almost  self-evident,  is  of  great  importance  in 
this  connection  as  it  enables  us  to  calculate  the  sine  of  a 
very  small  angle  with  a  minimum  of  effort. 

If  an  angle  or  arc  is  expressed  in  radians^  the  quotient  — ^— 

u 

approaches  the  limit  1  when  the  angle  itself  approaches  zero  as 

a  limit.     In  symbols 


I 


limit  ?i5L?=:l. 


194 


FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 


Fig.  104 


But 


In  order  to  prove  this  theorem,  let 
us  draw  an  acute  angle  A  OF  =  ^  as  in 
Fig.  104,  and  symmetrically  the  angle 
AOQ  also  equal  to  6.  With  the  vertex 
0  as  center  and  any  convenient  radius 
r,  draw  the  circular  arc  PAQ  and  join 
FQ,  intersecting  OA  in  M.  The  tan- 
gents FT  and  QT,  at  F  and  Q,  will 
meet  in  a  point  T  of  OA  prolonged. 

Obviously  we  shall  have 
(1)     FQ  <  arc  FAQ  <  FT+  TQ. 


FQ  =  2  FM=  2  r  sin  0, 
arc  FAQ  =  2aTGAF  =  2  rd    (Art.  74,  equation  (5)), 
FT+TQ=2FT^2ri2.nd, 

where  the  truth  of  the  second  equation  depends  essentially 
upon  our  assumption  that  6  is  expressed  in  radians.  If  these 
values  be  substituted  in  (1),  we  find 

2  r  sin  (9  <  2  r  ^  <  2  r  tan  ^, 

or,  after  division  by  the  positive  quantity  2  r, 

(2)  sin  ^  <  (9  <  tan  6, 

If  we  divide  all  three  members  of  this  inequality  by  the 
positive  number  sin  ^,  we  find 

e  1 


(3) 


1< 


sin  6      cos  0 


We  know  that  cos  6  approaches  the  limit  1  when  6  ap- 

proaches  zero.     Since  the  value  of  — — -^  according  to  (3), 

sin  0 

,  which  latter  quantity  itself  ap- 


lies  between  1  and 


cos  6 


proaches  1, 
(4) 


e 


sin  6 


must  also  have  1  as  its  limit.     That  is, 


limit     ^ 
0*0   sin  0 


=  1, 


CERTAIN   LIMIT   RELATIONS  195 

From  (3)  we  have  further 

^  ^  sin  ^  ^         /, 
1  >  —5—  >  cos  6, 
u 

so  that,  by  a  similar  argument,  we  find 
(^^  limit  ?HL?_i 

If  we  divide  all  members  pf  (2)  by  the  positive  quantity 
tan  ^,  we  find 

cos  e  <  — ^  <  1, 

tan  a 
so  that 

(Q^  limit     ^     =  limit  ^^^  ^  =  1 

^  ^  «=^o   tan  (9       «^o       0 

Since 

sin  (  —  ^)  _  —  sin  6  _  sin  ^ 

:^^e~~   -e  ~~r' 

tan  ( —  ^)  _  —  tan  6  _  tan  6 

equations  (4),  (5),  and  (6)  will  still  remain  true  if  6  ap- 
proaches zero  through  negative  instead  of  positive  values. 

These  formulae  have  many  important  applications.  For 
the  present,  we  shall  mention  only  the  one  indicated  at  the 
beginning  of  this  article.  The  content  of  equation  (5)  may 
be  formulated  as  follows.     If  we  write 

(T)  !i^=l-S, 

the  angle  6  (expressed  in  radians)  may  be  chosen  so  small 

that  h  i  the  difference  between  1  and  — ^—  j  will  become  less 

than  any  previously  assigned  quantity.  Since  we  find 
from  (7)  ^      ^  ^^ 

we  see  that  we  can  make  the  angle  6  (expressed  in  radians) 
so  small  that  the  difference  between  6  and  sin  6  becomes  less 
than  any  previously  assigned  small  fraction  of  0. 


196    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 

Suppose,  for  instance,  that  we  wish  to  compute  the  sine  of 
a  small  angle  to  5  decimal  places,  that  is,  with  an  error  which 
shall  be  less  than  5  units  of  the  sixth  decimal  place,  or 
.000005.  We  now  know  that  the  angle  may  be  chosen  so 
small  that  its  sine  may  be  equated  to  the  radian  measure  of 
the  angle  itself  with  an  error  of  less  than  5  units  of  the  sixth 
decimal  place.     In  other  words,  the  equation 

(8)  sin  6=0  (in  radians) 

will  be  true  up  to  five  decimal  places  for  all  angles  which  are 
sufficiently  small. 

Of  course,  our  method  does  not  inform  us  just  how  small 
6  must  be  in  order  that  equation  (8)  may  be  true  up  to  five 
decimal  places.  It  would  take  us  too  far  afield  to  investigate 
this  question,  a  complete  answer  to  which  is  beyond  the  scope 
of  this  book.  The  student  may  convince  himself,  however, 
by  actual  comparison  with  the  tables,  that  equation  (8)  is 
true  to  five  decimal  places  for  all  angles  less  than  2°.  In  all 
numerical  work,  then,  involving  such  small  angles,  no  error 
noticeable  in  five-place  calculations  is  introduced  by  putting 
sin  6=6  (in  radians). 

Since  we  have 

cos  ^  =  1  -  2  sin2-  (Art.  84,  equation  (1)), 

we  may  put 

(9)  cos^=l-2(|J=l-l^, 

a  formula  which  will  certainly  be  true  up  to  the  fifth  decimal 
place  for  all  angles  less  than  2°.  In  fact  equation  (9)  holds  to 
five  decimal  places  even  for  angles  much  larger  than  this  and 
may  serve  for  the  purpose  of  computing  the  cosines  of  such 
angles.  Since  6^  is  small  as  compared  with  6,  if  6  itself  is 
small,  we  shall  even  be  justified  in  equating  cos  ^  to  1  for 
very  small  angles.  Our  tables  show  that  no  error  is  intro- 
duced in  five-place  calculations  by  putting  cos  ^  =  1,  if  ^  is 
less  than  0°  16'. 


CERTAIN  LIMIT   RELATIONS  197 

Since  (8)  is  true  to  five  decimal  places  if  ^  <  2°,  we  see 
that  we  shall  have 

sin  26  =  26 

with  the  same  degree  of  approximation  if  ^  <  1°.  More 
generally  the  formula 

(10)  sin  n6  =  n6  (6  in  radians) 

is  correct  to  five  decimal  places  if  n6  is  less  than  2°. 

The  results  deduced  in  this  article  make  it  very  easy  to 
compute  the  functions  of  very  small  angles.  By  combining 
these  results  with  the  methods  of  Art.  83  an  extensive  table 
of  the  trigonometric  functions  may  be  constructed  with  com- 
parative ease. 

EXERCISE   L 

Compute  the  values  of  the  following  functions  of  small  angles  to  five 
decimal  places  by  the  method  of  Art.  85  a  and  compare  with  the  values 
obtained  from  the  table : 

1.  sin  12'.  3.   sin  1°.  5.   tan  1°. 

2.  tan  15'.  4.   cos  1°.  6.   cot  1°. 

7.  What  will  be  the  angle  subtended  by  a  lamp-post  10  feet  high  at  a 
distance  of  one  mile? 

8.  In  order  to  find  the  distance  from  the  earth  to  the  moon,  the 
following  plan  may  be  adopted.  Two  astronomers  stationed  at  A  and  B 
respectively  (Fig.   105)   observe 

at  the  same  instant  the  angular 
distance  of  the  moon's  center  M 
from  their  respective  zeniths 
(their  overhead  points),  Z  and 
Z'.     This  gives  the  angles 

a  =  Z^M  and  /3  =  Z'BM. 
For  the   sake  of    simplicity  as-  Fig.  105 

sume  that  both  stations  A   and 

B  are  on  the  equator,  that  the  moon  is  in  the  plane  of  the  equator,  and 
let  E  be  the  center  of  the  earth.  Then  /.  A  EB  =  X  is  equal  to  the  differ- 
ence between  the  longitudes  of  the  two  stations  and  may  be  regarded  as 
known.     We  may  now  compute 

^EAM=  180°  -a,  /.  EBM  =  180°  -  /3,  ZAEB  =  X. 


198    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 

Now  the  sum  of  the  four  angles  of  the  quadrilateral  AEBM  is  four  right 
angles ;  that  is, 

^M+  360°  -  a  -  y8  +  A.  =  360=, 
whence 

From  the  value  of  M  obtained  in  this  way,  it  is  easy  to  compute  the 
angle  subtended  by  the  earth's  radius  at  the  center  of  the  moon.  This 
angle  is  called  the  moon's  parallax. 

Find  the  moon's  distance  from  the  earth  if  the  moon's  parallax  is  57' 
and  if  the  earth's  radius  is  4000  miles. 

9.  The  apparent  diameter  of  the  moon  as  seen  from  the  earth  is 
about  31'.  Making  use  of  the  result  of  Ex.  8,  what  is  the  moon's 
diameter  in  miles? 

10.  The  sun's  parallax  is  about  8".8.  Assuming  4000  miles  as  the 
length  of  the  earth's  radius,  find  the  distance  from  the  earth  to  the  sun. 

85  b.  The  aixxiliary  quantities  S  and  T.  We  have  seen  in 
Art.  28  that  the  ordinary  tables  of  sines  and  tangents  be- 
come inconvenient  for  very  small  angles.  To  avoid  this  in- 
convenience, we  constructed  an  additional  table  (Table  III), 
giving  the  values  of  the  sines  and  tangents  of  such  small 
angles  directly  for  every  second  of  arc.  But  we  may  accom- 
plish the  same  purpose  in  another  way,  by  means  of  an 
auxiliary  table  occupying  far  less  space  than  the  additional 
table  just  mentioned.  This  second  method  is  based  on  the 
fact  that  the  quotients 

sin  0       J  tan  6 

change  very  slowly  if  ^  is  a  small  angle. 

We  have  just  seen  that  each  of  these  quotients  has  unity 
as  its  limit  when  6  approaches  zero,  provided  that  the  angle 
is  measured  in  radians.  Let  us  instead  express  6  in  minutes 
of  arc.  Let  6'  denote  the  number  of  minutes  and  B^^^  the 
number  of  radians 'contained  in  the  angle  6.  Then,  accord- 
ing to  Art.  85  a, 

(1)  limit  5yLi  =  limit  i5]2i=l. 


AUXILIARY   QUANTITIES   5  AND    T  199 

Since  (Art.  74) 

1°  =  -^  radians  =  0.0174533  radian, 

180 

and  therefore         y  ^  0.0002909  radian, 

the  angle  6^  which  contains  6'  minutes,  will  contain 

l9(^)  =0.000290919'  radian.     " 
Consequently  we  find 

sin  ^       .     n  0^^  n  nAAonr^n  sin  e 


=  sin  ^  -^  TT-^r^::^— —  =  0.0002909 


6'  '  0.0002909        *  ^(^  ' 

and  therefore,  on  account  of  (1), 

(2)  limit  !HL^  =  limit  i^  =  0.0002909. 

In  other  words,  if  6'  is  an  angle  expressed  in  minutes,  the 

common  limit  toward  which  and  — — —  tend,  when  6' 

6  6' 

approaches  zero,  is  a  number  whose  first  seven  decimal  places 

are  given  by  0.0002909. 

Let  us  write 

^Q\  sin  6  .      tan  6 

(3)  «  =  -^,         «  =  -^- 

These  quantities  change  their  values  very  slowly  for  small , 
values  of  0.     In  fact  we  have  just  seen  that,  for  angles  which 
are  small  enough,  we  shall  have 

log  8  =  \ogt  =  log  0.0002909  =  6.46373  -  10. 

For  ^  =  2°  =  120'  we  have 

^  sin  2°  ^  tan  2° 

^~   120  '  120   ' 

which  gives,  if  we  look  up  the  logarithms  from  the  tables, 

log  sin  2°  =  8.54282  -  10  log  tan  2°  =  8.54308  -  10 

log  120  =  2.07918  log  120  =  2.07918 

log  8  =  6.46364  -  10  log  t  =  6.46390  -  10 

Therefore,  while  0  changes  from  0°  to  2°,  log  s  changes  only 
by  9  units  and  log  t  by  17  units  of  the  fifth  decimal  place. 


200    FUNCTIONS  OF  MORE  THAN  ONE  ANGLE 

Table  IV  enables  us  to  find  the  values  of  iS=\ogs  and 
T=  log  t  for  every  angle  between  0°  and  2°.  To  find  the 
logarithm  of  the  sine  or  tangent  of  such  an  angle  we  have 
the  formulae  (immediate  consequences  of  (3)) 

.^.  log  sin  (9  =  log  5>' +  log  8  =  lojf  ^' +  aS', 

^  ^  log  tan  0  =  log  6'  +  log  ^  =  log  6'  +  T, 

If  the  log  sin  or  log  tan  of  a  small  angle  is  given,  to  find 
the  angle,  we  may  do  this  by  means  of  one  of  the  equations 

.rN  log  6^  =  log  sin  6  —  S, 

^  ^  log  6'  =  log  tan  0  -  T, 

obtained  from  (4)  by  transposition. 

Of  course  the  quantities  aS'  and  T  are  available,  not  only 
for  sines  and  tangents  of  small  angles,  but  also  for  cosines 
and  cotangents  of  angles  close  to  90°. 

EXERCISE    LI 

1.  Find  the  sine  and  tangent  of  1°  13'.21  by  using  the  auxiliaries  S 
and  T. 

Solution.     Since  6=1°  13'.21,  we  have  0'  =  73'.21. 

log  0'  =  1.86457  log  6'  =  1.86457 

S  =  6.46369  -  10  7^^6.46379 -10 

log  sin  d  =  8.32826  -  10  log  tan  ^  =  8.32836  -  10 

2.  Given  log  sin  9  =  8.24798  -  10.     Find  6. 

Solution.     We  find  from  Table  IV  cori:esponding  to  log  sin  ^=8.24798 
-  10,  5  =  6.46370.     Formula  (5)  leads  to  the  calculation 
log  sin  0  =  8.24798  -  10 
S  =  6.46370  -  10 
log  6'  =  1.78428  .'.6'  =  60'.85  =  1°  0'.85. 

Find  the  values  of  the  logarithms  of  the  following  functions  by  means 
of  the  auxiliaries  S  and  T:  I 

3.  sinl°21'.63.  5.   cos  89°  13'.21.  | 

4.  tan0°32'.61.  6.  cot  88°  21'.75. 

Find  the  angles  determined  by  the  following  functions  by  means  of 
S  and  T: 

7.  log  sin  $  =  7.76345  -  10.  9.   log  cos  0  =  8.42371  -  10. 

8.  log  tan  0  =  8.50731  -  10.  10.   log  cot  0  =  8.53729  -  10. 


■r^ 


CHAPTER   XII 

DIRECTED   LINES   AND   DIRECTED  LINE-SEGMENTS  * 

86.  Plan  of  another  proof  for  the  addition  formulae.  When 
we  proved  the  addition  theorem  in  Art.  79,  we  found  it 
necessary  to  divide  the  proof  into  a  number  of  cases  accord- 
ing as  the  angles  were  in  the  first,  second,  third,  or  fourth 
quadrants.  To  be  sure,  by  making  use  of  the  method  of 
mathematical  induction  we  found  it  a  fairly  simple  matter  to 
make  an  exhaustive  discussion  covering  all  cases.  Neverthe- 
less we  feel  that  it  must  be  possible  to  devise  a  method 
enabling  us  to  prove  this  theorem  at  one  stroke  for  angles  of 
any  magnitude.  The  key  to  the  solution  of  this  problem  is 
found  to  be  a  careful  formulation  of  the  notions  of  a  directed 
line  and  a  directed  line- segment.  Since  these  notions  are  of 
very  great  importance,  not  only  in  this  connection,  but  in 
many  other  parts  of  pure  and  applied  mathematics,  we  shall 
find  it  worth  our  while  to  speak  of  them,  even  if  they  are 
not  absolutely  indispensable  for  the  proof  of  the  addition 
theorem. 

87.  Directed  lines  and  segments.  A  straight  line  is  infi- 
nite in  extent  and  is  determined  by  any  two  distinct  points 
upon  it.  We  may,  however,  think  of  one  and  the  same 
straight  line  as  having  either  of  two  opposite  directions,  in 
which  case  we  speak  of  it  as  a  directed  line.  Since  we  can 
never  draw  more  than  a  finite  portion  of  a  line,  we  may 
indicate  the  direction  of  a  directed  line  by  placing  a  -h  sign 
near  one  end  of  that  portion  which  actually  appears  in  the 
figure.     In  Fig.  106  we  have  thus  indicated  the  direction  of 

*  This  chapter  may  be  omitted  in  a  first  course  if  the  time  is  insufficient. 

201 


202 


DIRECTED  LINES   AND   LINE-SEGMENTS 


the  directed  line  I  which  is  to  be  thought  of  as  pointing 
toward  the  upper  right-hand  corner  of  the  page. 

This  method  of  indicating  the  direction  of  a  di- 
rected line  has  ah'eady  been  used  in  this  book  to 
indicate  the  positive  direction  of  the  x-axis  and  y-axis 
of  a  system  of  rectangular  coordinates.  (See  Art.  63.) 
These  are  directed  lines. 

A  line -segment  is  a  finite  portion  of  a  line 
and  may  be  described  by  naming  its  end  points,  such  as  AB 
in  Fig.  106.  But  again,  we  may  think  of  it  as  a  directed 
line-segment,  thus  distinguishing  between  AB  and  BA. 

When  a  directed  line-segment  lies  upon  a  directed  line,  its 
direction  or  sense  may  be  the  same  as  that  of  the  line  or  else 
opposite  to  it.  If  a  directed  line-segment  on  Z  is  5  units 
long  and  if  its  djrectioil  is  the  same  as  that  of  Z,  we  may 
represent  it  by  the  number  +5.  A  line-segment  of  the 
same  length  and  opposite  direction  will  be  represented  by  —  5. 
In  general,  a  directed  line-segment,  which  lies  on  a  directed 
line,  shall  be  counted  positive  or  negative  according  as  it  has  the 
same  or  the  opposite  direction  as  the  directed  line.  For  such 
line-segments,  we  always  have  BA  =  —AB,  or  AB  +  BA  =  0. 

We  are  now  ready  to  prove  the  following  theorem.  If  A, 
B,  C,  are  any  three  points  on  a  directed  line,  then 

(1)  AB  +  BC=AC, 

where  AB,  BC,  and  AC  are  directed  line-segments. 

Proof.  1.  Let  J. (7  be  positive  and  let  B 
be  between  A  and  O.  Then  AB  and  BC  are 
also  positive  and  the  truth  of  the  theorem,  in 
this  case,  is  obvious.     (See  Fig.  107.) 

2.  Let   AC  hQ   positive,  but  let  C  be   be- 
tween A  and  B.     Then  AB  is  positive,  but  BC 
is  negative  and  equal  to  —  CB.     Thus 
AB  +  BC=AB-CB  =  AC      (See  Fig.  108.) 

3.  Let  AC  still  be  positive,  but  let  A  be  be- 
tween B  and  C     Then  (Fig.  109), 

AB  -f  BC=  -BA^-  BC=  BC-BA  =  AC.  fiq.  io9 


Fig.  107 


BX 


Fig.  108 


C^* 


DIRECTED  LINES  AND   SEGMENTS  203 

Ji  AOis  negative,  there  are  again  three  cases  according  as 
the  order  of  the  three  points  is  CBA^  BOA,  or  CAB.  But 
the  above  relation  (1)  will  be  found  to  be  true  in  all  cases. 
These  last  three  cases  may,  of  course,  also  be  reduced  to  the 
former  three  by  reversing  the  direction  of  the  line  I. 

The  following  is  a  simple  corollary  of  the  above  theorem, 
if  A,  B,  C,  i>,  are  any  four  points  of  a  directed  line^  we  have 
the  relation 

(2)  AB  +  BC+CD:=AD 

between  the  directed  line-segments  AB^  BO^  CD^  and  AD* 

In  fact,  by  the  theorem  just  proved,  we  have 

AB  +  BQ^AC,  AC-hCI)  =  AD, 

so  that  we  find  by  addition 

AB  +  BC+  AC -\-  CD  =  AC -\-  AD, 

which  reduces  to  (2)  if  we  subtract  A  (7  from  both  members. 

It  may  now  be  proved  by  induction  that,  in  general,  if 
A,  B,  (7,  •••  M,  I^  are  any  finite  number  of  points  on  a  directed 
line,  then 

(3)  AB-hBC+CD-{-  ...  -\-MJS'=AN. 

Equations  (1),  (2),  (3)  may  also  be  proved  by  algebra. 
On  the  directed  line  I,  let  us  introduce  a  point  0  as  origin 
or  zero  point  of  a  scale,  whose  positive  readings  are  on  that 
side  of  0  which  corresponds  to  the  positive  direction  of  the 
line  I.  This  is  precisely  what  we  did  when  we  established 
scales  upon  the  a:-axis  and  ?/-axis  of  a 
coordinate  system  (Art.  63).  Denote 
by  Ij^  the  reading  of  the  scale  which 
corresponds  to  the  point  A,  by  l^  that 
which  corresponds  to  B.  The  differ- 
ence Iff  —  Ij^  will  give  the  length  of  the  line-segment  AB, 
affected  with  a  plus  or  minus  sign  according  as  AB  is  a 
positive  or  negative  line-segment  in  the  sense  of  our  pre- 
vious definition. 


204  DIRECTED  LINES   AND  LINE-SEGMENTS 

If  then  we  have  any  three  points  A,  B,  Q  on  the  directed 
line,  we  shall  have 

AB^l^-h.         BQ^la-lB.        AO=la-h, 
and  therefore 

AB  +  BC=  Ib-Ia  +  Ic-Ib=Ic-Ia=-AO, 
which  is  the  same  as  (1).     In  the  same  way  we  may  also 
prove  equations  (2)  and  (3). 

88.  Angles  between  directed  lines.     Let  I  and  m  be  two 

directed  lines,  and  let  us  denote  the  angle  between  their 
positive  directions  by  (Z,  m)  or  (w,  I)  according  as  we  think 
of  Z  or  m  as  the  initial  side  of  the  angle.  To  be  perfectly 
specific,  we  understand  by  (Z,  m)  the  angle,  less  than  360°, 
through  which  it  would  be  necessary  to 
rotate  the  directed  line  I  in  the  counter- 
clockwise direction,  in  order  to  make  its 
positive  direction  coincide  with  the  posi- 
tive direction  of  the  directed  line  m.  In 
Fig.  Ill  Fig.  Ill,  the  angle  (I,  m)  is  marked  d. 

Similarly    (m,  V)  is  the   angle   through 
which  m  would  have  to  be  turned  in  the  positive  (counter- 
clockwise) direction  in  order  to  make  the  positive  direction 
of  m  coincide  with  that  of  I.     In  Fig.  Ill  (m,  V)  is  marked  6' . 
We  see  that  we  shall  always  have 

(1)  0,  m)  +  (w,  Z)=360°, 
so  that 

(m,  0=360°-  (I,  m) 

and  therefore  (see  Art.  77,  equations  (7)), 

(2)  sin  (m,  V)=  —  sin  (?,  w),         cos  (w,  Z)  =  cos  (Z,  m). 

89.  Projections.  The  projection  of  a  point  P  on  a  line  Z 
is  the  foot  of  the  perpendicular  dropped  from  the  point  to 
the    line.       The    projection    of    a    line-segment 

AB  on  a  line  Z  (see  Fig.  112)  is  the  line-seg- 
ment A B'  of  Z  bounded  by  the  projections  of 
A  and  B.  If  AB  is  a  directed  line-segment,  so 
is  A'B'. 


-' 

i^^ 

A 
I 

'         1 
"lO.  11 

3> 

2 

PROJECTIONS 


205 


We  wish  to  solve  the  following  problem.  G-iven  a  directed 
line-segment  AB  on  a  directed  line  p ;  to  find  the  magnitude 
and  sign  of  its  projection  upon  any  second  directed  line  I. 

The  line-segment  AB  may  have  the  same  sense  as  the 
directed  line  p  or  else  the  opposite  sense ;  it  will  be  repre- 
sented by  a  positive  number  in  the  first  case  and  by  a  nega- 
tive number  in  the  second  (Art.  87).  If  we  denote  by 
I  AB  I  the  positive  number  which  represents  merely  the 
length  (regardless  of  direction)  of  the  line-segment  AB, 
we  shall  have 

(1)  AB  =  \AB  |,  read  AB  =  length  AB, 
or 

(2)  AB  =  —  I  AB  |,  read  AB  =  minus  length  AB, 
according  as  the  direction  of  AB  agrees  with  that  of  the 
directed  line  p  or  not. 

Let  us  consider  first  the  case  (1)  in  which  AB  is  positive. 
(See  Figs.  113  and  114.)  Let  OM  be  the  projection  of  AB 
on  L     Choose  0  as  origin  and  I  as  the  a;-axis  of  a  system  of 


^^ 


Fig.  113 


Fig.  114 


O^ 


coordinates,  so  that  the  positive  a;-axis  coincides  with  the 
positive  direction  of  the  line  I.  The  line  m  through  0,  per- 
pendicular to  I  and  with  its  positive  direction  as  indicated  in 
the  figures,  will  then  be  the  «/-axis.  Through  0  draw  a 
directed  line  p'  parallel  to  p  and  let  P  be  its  intersection 
with  the  line  MB  through  B  perpendicular  to  I*     The  line- 

*  The  word  parallel  in  this  theory  means,  not  only  that  the  lines  are  parallel 
in  the  ordinary  sense,  but  that  their  positive  directions  are  the  same.  The  word 
anti-parallel  is  sometimes  used  for  two  parallel  directed  lines  whose  positive' 
directions  are  opposite. 


206 


DIRECTED  LINES  AND  LINE-SEGMENTS 


segments  AB  and  OP  have  the  same  length  and  direction, 

and  the  same  directed  line-segment  OM  on  I  as  projection. 

But,   by   the   definition  of  the   cosine  of   a   general   angle 

(Art.  64),  OM 

cos  (I,  p)  =  cos  (I,  p')  =  — , 

since  OMis  the  abscissa  and  OP  the  radius  vector  of  a  point 
P  on  the  terminal  side  of  the  angle  (Z,  p'^,  this  angle  being 
in  its  standard  position  and  OP  =  AB  being  positive  in  the 
case  under  consideration.     Consequently  we  have 

0M=  OP  cos  (Z,  p)  =  AB  cos  (Z,  p). 
Since  OM  is  the  projection  of  AB  on  Z,  we  may  write  this 
as  follows : 

(3)     0M=  proji  AB  =  AB  cos  (Z,  p)  =  AB  cos  (jd,  Z)  ;* 
for,  according  to  equation  (2),  Art.  88,  the  angles  (Z,  p)  and 
(^,  Z)  have  the  same  cosine. 

Thus  the  projection  upon  the  directed  line  Z,  of  the  positive 
line-segment  AB  of  the  directed  line  p^  is  equal  to  AB  mul- 
tiplied by  the  cosine  of  the  angle  between  Z  and  p.  The 
projection  is  a  directed  line-segment  on  Z,  positive  if  (Z,  p}  is 
p  in  the  first  or  fourth,  negative  if 
(Z,  jt?)  is  in  the  second  or  third 
quadrant. 

We  have  proved  equation  (3) 
when  AB  is  positive.  Let  us  con- 
sider now  the  case  when  AB  is  a 
negative  line-segment  of  p,  (See 
Fig.  115.)  The  projection  oiAB, 
as  well  as  that  of  OP,  is  OM. 
We  may  think  of  OP,  which  is  a  negative  line-segment  on 
p\  as  a  positive  line-segment  of  a  directed  line  p'^,  coincident 
with  p'  but  having  the  opposite  direction.  But  this  latter 
directed  line  makes  an  angle  with  Z  just  180°  greater  than 
(Z,  p')  ;  that  is. 


+m 


Fig.  115 


(4) 


<il,p")  =  il,p'}  + ISO' 


♦  The  symbol  proj^  AB  is  read  "projection  of  AB  upon  l." 


PROJECTION  OF   A  BROKEN  LINE  207 

For  if  we  rotate  I  in  the  counterclockwise  direction  around 
0  as  center,  it  will  take  180°  more  to  make  +  I  coincide  with 
4-  p"  than  with  +  p' .  According  to  formula  (3),  which  we 
know  to  be  valid  for  all  positive  line-segments,  we  have 
therefore  ^^^.^  AB=\AB\  cos  (I,  p"}, 

where  |  AB  \  denotes  the  length  of  the  line-segment  AB  taken 
as  a  positive  number.  But  according  to  (4)  and  equations 
(4)  of  Art.  77,  we  have 

cos  (I,  p"^  =  —  cos  (I,  p')  =  —  cos  (I,  p}  =—  cos  (jo,  ?), 

so  that 

proji  AB  =  —  I  AB  \  cos  (I,  jo)  =  —  |  AB  \  cos  (p,  ?). 

Since  in  our  case  AB  was  a  negative  line-segment,  we  had 
(cf.  equation  (2)) 

AB  =  -\AB\, 

so  that  we  may  write  finally 

(3)  proji  AB  =  AB  cos  (Z,  p)  =  AB  cos  (p,  I). 

In  other  words,  formula  (fi)  for  the  projection  upon  a  directed 
line  l^  of  a  directed  line-segment  AB  of  a  second  directed  line 
p,  is  true  for  both  positive  and  negative  line- segments. 

If  we  think  of  the  line-segments  as  mere  lengths,  not 
endowed  with  direction,  formula  (3)  may  be  simplified  to 

A'B'  =  proji  AB  =  AB  cos  6,  r^      ^ 

where  6  is  the  angle  between  AB        ^^<9r  r T^y^ 

and  the  line  L  this  an^le  being:  un-  ^  ^  ^ 

,        ,       ,  .         '  ^„    -         ^,  Fig.  116  Fig.  117 

derstood  in  the  sense  oi  elementary 

geometry  without  any  reference  to  a  positive  sense  of  rota- 
tion. Consequently  6  may  be  regarded  as  an  acute  angle, 
so  that  its  cosine  will  always  be  positive.  (See  Figs.  116 
and  117.) 

90.   Projection  of  a  broken  line.     Let  us  connect  two  points 
A  and  (7  by  a  directed  line-segment  AC.     We  may  think  of 


208 


DIRECTED  LINES  AND  LINE-SEGMENTS 


this  line-segment  as  describing  the  shortest  path  from  A  to 
Q  in  magnitude  and  direction.  Let  ABC  be  a  second 
B  (longer)  path  from  A  X^o  C  made  up  of 

two  directed  line-segments  AB  and  BQ 
(Fig.  118). 

Let  us  project  these  three  line-segments 


B'       c 
Fig.  118 


-f-l 


on  any  directed  line  ?,  so  that 

(1)  A'B^  =  proj,  AB,  B'  Q'  =  proj^  BC,  A'  C  =  proj^  AO. 

Since  A'B',  B' Q'  and  A' C  are  three  directed  line-segments 
of  a  directed  line,  we  have  (Art.  87,  equation  (1)) 

A'C^  =A'B' +  B'0'l 

whence,  substituting  the  values  (1), 

(2)  proj,  ^(7=proji  Ji^  +  proji^a 

By  means  of  the  more  general  relation  (3)  of  Art.  87,  we 
may  prove  the  following  general  theorem,  of  which  (2)  ex- 
presses the  simplest  special  case. 

Let  us  connect  two  points  hy  two  paths,  one  of  which  is  a 
single  directed  line-segment,  while  the  other  is  made  up  of  a 
finite  number  of  directed  line-segments.  Then  the  projection, 
upon  any  directed  line,  of  the  first  path  is  equal  to  the  sum  of 
the  projections  of  the  various 
line-segments  which  make  up  J^ 

the  second  path.  ^  wi^^O^     e^ 

Figure  119,  in  which  the  va- 
rious line -segments  are  marked 
with  arrowheads  to  indicate  their 
directions,  illustrates  this  theo- 
rem. Since  the  positive  direc- 
tion of  I  is  toward  the  right,  the  projections  A^G^,  A'B',  B'C,  D'E', 
E'F',  F'G'  are  all  positive,  while  the  projection  CD'  of  CD  is  negative. 
Consequently 

proj,  AB  +  projz  BC  -\-  proj,  CD 

=  A'B'  +  B'C  +  CD'  =  A'B'  -\-  B'C  -  D'C  =  A'D'- 

Let  the  line  I  in  Fig.  118  coincide  with  AC,  and  denote  by 
a,  6,  c  the  lengths  of  the  three  line-segments  BC,  AC,  and 


B'  D'    C       E' 
Fia.  119 


DIRECTION  COSINES   OF   A  LINE 


209 


AB  respectively.      (See   Fig.  120.) 
According  to  (2)  and  Art.  89,  equa- 
tion (3),  we  shall  have 
(3)        b  =  c  cos  A-\-  a  cos  (7, 
a  relation  which  may  be  used  to  ad- 
vantage in  many  problems  concerning  triangles 
the  two  equations  similar  to  (3) 

e=  a  cos  B  -\-h  cos  A^ 
a  =  b  cos  C-\-  c  cos  B, 


b         c 

Fig.  120 


■¥l 


Of  course 


(4) 


may  be  proved  in  the  same  fashion. 

91.  Direction  cosines  of  a  line.  Let  V  be  a  directed  line 
through  the  origin  of  coordinates.  The  angles  (a;,  V}  and 
(Z',  ?/),  which  its  positive  direction  makes  with  the  positive 
2;-axis  and  y-axis,  are  called  its  direction  angles.  The  direc- 
tion angles  (x^  V)  and  (Z,  y)  of  a  line  I  which  does  not  pass 
through  the  origin  are  defined  to  be  the  same  as  those  of  a 
parallel  line  V  which  does  pass  through  the  origin.  Observe 
that  the  angle  (x^  I)  has  the  2;-axis  as  initial  side,  while  the 
initial  side  of  the  second  direction  angle  (Z,  y)  is  not  the 
y-axis,  but  the  line  I. 

If  the  angle  (a:,  Z)  is  in  the  first  quadrant  (Fig.  121),  we 
clearly  have 
(1)  (X,  O  +  (Z,y)=90°. 


+« 


If  the  angle  (x^  V)  is  in  the  second  quadrant  (Fig.  122), 
(l,y)  =  (l',y-)=mO°-a. 


210 


DIRECTED  LINES  AND  LINE-SEGMENTS 


For  clearly  it  requires  a  positive  (counterclockwise)  rotation 
of  360°  —  a  to  make  +  V  coincide  with  -|-  y.  Therefore  we 
have,  in  this  case, 

(2)  (2:,  O+fty)=360°  +  90°, 

and  the  same  relation  holds  if  (x^  V)  is  in  the  third  or  fourth 
quadrant,  as  may  be  verified  easily.  Thus  we  always  have 
either  (Z,  ^)=90°-(2;,  Z)  or  (Z,  ^)  =  360°  +  90°  -  (a:,  0,  so 
that  in  all  cases 

(3)  cos  (y,  I)  =  cos  (Z,  ?/)  =  sin  (2;,  Z),  sin  (Z,  y)  =  cos  (2:,  I). 
The  two  quantities  cos  (2;,  Z)  and  cos(^,  Z)  are  called  the 
direction  cosines  of  the  directed  line  Z. 

Since  we  have,  for  any  angle, 

cos2  (2:,  Z)  H-  sin2  (2;,  Z)  =  1, 
we  find  from   (3)  the  following  simple  relation  between  the 
direction  cosines  of  a  line  I ; 
(2)  cos2  (x,  I)  +  cos2  (y,  I)  =  1. 

92.  Formula  for  the  cosine  of  the  angle  between  two  lines 
whose  direction  cosines  are  given.  Let  us  consider  two 
directed  lines  Z  and  m  whose  direction  cosines  are  cos  (2^,  Z), 
cos(y,  Z),  and  cos  {x,  m),  cos(y,  rn)  respectively.  We  wish 
to  find  a  formula  for  the  cosine  of  the  angle  between  the 
two  lines. 

We  may  assume  that  the  lines  Z  and  m  pass  through  the 
origin.     If  they  do  not,  we  may  first  solve  the  problem  for 

two  lines  Z^  m',  parallel  to  Z,  m  re- 
spectively, which  do  pass  through 
the  origin.  Since  Z^  mJ  have  the 
same  direction  cosines  as  Z  and  w, 
and  since  the  angle  between  l\  m'  is 
equal  to  that  between  Z,  w,  the  two 
problems  are  clearly  equivalent. 

Let  us  choose  a  point  M  (see 
Fig.  123),  on  the  line  w,  such  that 
the  line-segment  OM  is  positive.  Let  X  be  the  projection 
of  M  on  the   a;-axis.     Then  the  projection  of  the  broken 


+ 

y 

+ 

y' 

/ 

/ 

^ 

0' 

X 

Fig.  123 


COSINE  OF  ANGLE   BETWEEN   TWO  LINES         211 

line  OXM  on  I  will  be  equal  to  that  of  OM  (Art.  90,  equa- 
tion (2)).     That  is, 

(1)  proj^  0M=  proji  OX  +  proj^  XM. 

On  account  of  Art.  89,  equation  (3),  we  have 
proji  0M=^  OM  cos  (w,  Z), 
^^)     proji  OX  =  OX  cos  Cx,  0,     proji  XM=^  XMcos  (y,  Z), 
since  XM  is  a  directed  line-segment  on  a  directed  line  y 
parallel  to  the  ?/-axis.* 

Substitution  of  (2)  in  (1)  gives 

(3)  Oi!fcos(m,  0  =  OX  cos  (a:,  l)+XMcos{i/,  Z). 
But  ox=  ^Yoj^OM=  OM  cos  (x,  w), 

-Of  =  projy/Oif  =  i^ToiyOM=  OiHf  cos  (i/,  w), 

since  the  projections   of    OM^  on  the  two  parallel  directed 
lines  ^  and  y,  are  equal.     If  we  substitute  these  values  in 
(3),  we  find 
OiHf  cos  (w,  /) 

=  OM  cos  (a;,  Z)  cos  (a;,  w)  +  OM  cos  (^,  Z)  cos  (^,  w) 
or,  upon  division  by  OiHf, 

(4)  cos  (w*,  i)  =  cos  (a?,  Z)  cos  (a?,  m)  +  cos  (|/,  ?)  cos  (y,  ^, 
the  desired  formula. 

The  proof  which  we  have  given  of  this  formula  is  perfectly- 
general;  that  is,  it  is  applicable,  no  matter  how  large  or 
small  the  angles  (Z,  m),  {x,  Z),  etc.,  may  be,  in  what  quadrants 
they  happen  to  lie,  or  whether  they  are  positive  or  negative. 
In  fact,  the  figure  (Fig.  123)  has  not  really  been  used  in  the 
demonstration  except  for  the  purpose  of  suggesting  the  order 
of  the  various  steps  of  the  argument.  Every  step  of  this 
proof  can  be  justified  by  quoting  a  previously  demonstrated 
general  theorem. 

It  will  be  a  good  exercise  for  the  student  to  repeat  the  argument  with 
a  different  figure  in  which  some  or  all  of  the  angles  concerned  are  not 
acute. 

*  In  Fig.  123  we  have  chosen  the  positive  direction  of  y'  to  correspond  to  that 
of  y.    This  is  not  essential,  but  it  is  convenient. 


212  DIRECTED  LINES  AND   LINE-SEGMENTS 

93.  New  proof  for  the   addition  and  subtraction  formulae. 

Let  us  denote  the  angles  (x^  l)  and  (x^  m)  by  a  and  ff 
respectively.     Then 

cos  (?,  m)  =  cos  (a  —  yS), 
and  (Art.  91,  equations  (3)), 

cos  («/,  0  =  sin  (a:,  I)  =  sin  a,      cos  (?/,  w)  =  sin  (a;,  m)  =  sin  /S. 
Consequently  we  find,  from  equation  (4)  of  Art.  92, 
(1)  cos  (a  —  y3)  =  cos  a  cos  y8  -|-  sin  a  sin  y(3, 

the  subtraction  formula  for  the  cosine.  We  shall  leave  it  as 
an  exercise  for  the  student  to  deduce  from  this  the  subtrac- 
tion formula  for  the  sine  and  the  addition  formulae  for  both 
functions.     (See  Exercise  LIT.) 

94.  The  generalized  law  of  sines.  If  we  attribute  a 
definite  direction  to  every  line  of  the  plane,  and  define  angles 
between  them,  as  in  Art.  88,  with  reference  to  a  positive 
direction  of  rotation,  the  angle  between  any  two  such  directed 
lines  may  be  greater  than  180°  and  the  trigonometric  func- 
tions of  such  angles  will  have  definite  signs  as  well  as  numer- 
ical magnitude.  Moreover,  every  line-segment  will  then  also 
have  a  definite  sign. 

It  may  he  shown  that  the  law  of  sines,  when  written  in  the 
form 

sin  (6,  c)      sin  (<?,  a)      sin  (a,  6) 

will  he  true  of  any  triangle,  not  merely  with  regard  to  the  mag- 
nitudes of  the  quantities  involved,  hut  also  with  regard  to  their 
signs. 

In  order  to  explain  this  statement,  con- 
sider first  the  case  that  the  positive  direc- 
tions of  the  three  directed  lines  a,  h,  c 
(the  sides  of  the  triangle)  are  from  B 
toward  C,  from  C  toward  A,  and  from 
A  toward  B  respectively.  (Cf .  Fig.  124.) 
Fia.  124  Then  BO,  CA,  and  AB  are  all  positive, 


GENERALIZED  LAW  OF   SINES  213 

and  the  three  denominators  in  (1)  are  either  all  positive  or 
all  negative. 

In  Fig.  124  they  are  all  positive.  But  they  would  all  be  negative  if 
the  clockwise  rotation  had  been  chosen  as  positive  direction  of  rotation. 
They  would  all  be  negative,  even  without  any  change  in  the  choice  of  the 
positive  direction  of  rotation,  if  the  points  named  A  and  B  in  Fig.  124 
were  interchanged. 

Consequently  equations  (1)  are  true  in  this  case,  not 
merely  numerically,  but  also  with  regard  to  sign. 

Let  us  now  invert  the  positive  direction  of  a  single  one  of 
the  lines,  that  of  a,  for  instance.  Then  BO  becomes  negative; 
the  angle  (6,  c)  and  the  segments  OA  and  AB  remain  unal- 
tered ;  ((?,  a)  and  (a,  6)  each  change  by  180°,  so  that  their  sines 
change  sign.    Consequently  equations  (1)  will  still  be  verified. 

By  combining  several  such  changes  we  easily  arrive  at  the 
conclusion  that  equations  (1)  will  always  be  true,  in  magni- 
tude and  sign,  no  matter  how  the  positive  directions  of  the 
three  lines  a,  b,  c  may  have  been  selected  or  which  of  the  two 
opposite  kinds  of  rotation  be  regarded  as  positive. 

This  generalization  of  the  law  of  sines  is  due  to  the  great  geometer 
MoBius  (1790-1868),  and  is  of  great  importance  in  many  applications, 
especially  in  projective  geometry. 

EXERCISE    Lll 

1.  Show  that  the  principal  theorem  of  Art.  90  may  be  enunciated  as 
follows.  If  a  finite  number  of  directed  line-segments  form  a  closed  poly- 
gon, the  sum  of  their  projections  upon  any  directed  line  is  equal  to  zero. 

2.  Show  that  the  rc-component  of  the  resultant  of  two  forces  (cf. 
Art.  58)  is  equal  to  the  sum  of  the  x-components  of  the  two  original 
forces.     Similarly  for  the  y-components. 

3.  From  formula  (1)  of  Art.  93  deduce  the  formula  for  sin  («  —  yS). 

4.  From  formula  (1)  of  Art.  93  deduce  the  formulae  for  cos  («  4-  (3) 
and  sin  (a  -f  /?). 

5.  Generalize  the  law  of  tangents  from  the  standpoint  of  directed 
line-segments  in  a  way  which  shall  be  analogous  to  the  generalization  of 
the  law  of  sines  carried  out  in  Art.  94. 

Hint.     The  law  of  tangents  may  be  deduced  from  the  law  of  sines. 

6.  Generalize  in  similar  fashion  the  projection  formulae  (3)  and  (4) 
of  Art.  90. 


CHAPTER   XIII 

THE  INVERSE   TRIGONOMETRIC   FUNCTIONS   AND 
TRIGONOMETRIC   EQUATIONS 

95.  The  problem  of  inverting  the  trigonometric  fimctions. 

In  the  light  of  our  recent  studies  we  may  say,  with  a  con- 
siderable degree  of  propriety,  that  trigonometry  is  a  discus- 
sion of  the  properties  of  the  trigonometric  functions.  However, 
our  discussion  of  these  functions,  so  far,  has  been  somewhat 
one-sided.  With  the  exception  of  a  few  practical  applica- 
tions in  the  first  part  of  the  book,  we  have  always  looked 
upon  these  functions  from  the  point  of  view  of  what  may 
be  called  the  direct  problem ;  that  is,  given  the  angle,  to  find 
the  function.  We  now  propose  to  discuss,  somewhat  more 
fully  than  has  been  done  so  far,  the  inverse  problem  ;  that  is, 
given  the  value  of  one  of  the  functions,  to  find  the  corre- 
sponding angles. 

We  notice  at  once  a  fundamental  difference  between  these 
two  problems,  which  may  be  illustrated  by  the  relation  be- 
tween an  angle  and  its  sine.  Every  angle  has  only  one  sine, 
but  there  are  many  angles  which  have  the  same  sine.  Con- 
sequently while  the  direct  problem  (to  find  the  sine  of  a  given 
angle)  has  only  one  solution^  the  inverse  problem  (to  find  an 
angle  with  a  given  sine)  has  many  solutions. 

96.  Determination  of  all  of  the  angles  which  correspond  to 
a  given  value  of  one  of  the  functions.  When  we  are  solving 
triangles,  the  angles  are  necessarily  either  in  the  first  or  second 
quadrant,  so  that  in  most  cases  we  experience  little  difficulty 
in  finding  a  unique  angle  as  an  answer  to  a  given  problem 
of  this  kind.  But,  even  in  this  restricted  field,  we  found 
that  a  problem  may  have  two  solutions,  owing  to  the  fact  that 

214 


FINDING  ALL  ANGLES  FOR  A  GIVEN  FUNCTION      215 

there  exists  two  angles  ^,  one  acute  and  one  obtuse,  corre- 
sponding to  a  given  positive  value  of  sin  0.     (Cf .  Art.  bb.) 

If  the  sine  of  an  angle  is  given,  and  no  particular  quad- 
rant or  set  of  quadrants  is  prescribed  for  the  angle,  the 
number  of  values  which  the  angle  may  have  is  unlimited. 

We  may  prove  this  and  find  out  how  all  of  these  angles 
are  related  to  each  other  as  follows: 

Let  the  given  value  of  the  sine  be  denoted  by  «.  If  «  is 
numerically  greater  than  unity,  the  problem  has  no  solution, 
since  no  angle  has  a  sine  numerically  greater  than  1.  If  «  is 
numerically  less  than  1  and  positive,  there  exists  an  acute 
angle  0  and  an  obtuse  angle  180°  —  ^,  such  that 

sin  (9  =  sin  (180°-^)  =8. 

But,  on  account  of  the  periodic  character  of  the  sine,  we  also 

^^^^  sin  ((9  +  71 .  360°)  =  sin  (9  =  s, 

sin  (180°  -  6>  +  71  .  360°)  =  sin  (180°  -  ^)  =  s, 

where  n  is  any  positive  or  negative  integer  or  zero.  We  see 
therefore  that  all  angles,  such  as 

(1)  71 .  360°  +  (9  =  2  71 .  180°  +  <9, 
or 

(2)  n  .  360°  +  180°  -  ^  =  (2  7i  +  1)  •  180°  -  6, 

have  the  same  sine  as  the  angle  6.  This  may  be  expressed  as 
follows.  All  of  those  angles  which  can  be  obtained  by  add- 
ing a  given  angle  to  any  even  multiple  of  180°  or  else  by 
subtracting  the  given  angle  from  any  odd  multiple  of  180°, 
have  the  same  sine. 

That  these  are  the  only  angles  which  have  the  same  sine 
follows  easily  from  the  fact  that  two  distinct  angles  in  the 
same  quadrant  cannot  have  the  same  sine. 

If  the  given  value  of  8  is  negative,  nothing  essential  is 
changed  in  the  above  argument,  except  that  0  will  then  be  in 
the  third  or  fourth  quadrant  instead  of  being  an  acute 
angle.  It  will  still  be  true  that  all  of  the  angles  given  by 
formulae  (1)  and  (2),  and  no  others,  have  the  same  sine  as  6. 


216  INVERSE   TKIGONOMETRIC   FUNCTIONS 

Now  we  may  include  all  of  the  angles  (1)  and  (2)  in  the 
single  expression 

(3)  m  .  180°  ±  e, 

where  m  may  be  any  integer,  even  or  odd,  and  where  the  + 
or  —  sign  is  to  be  used  according  as  m  is  even  or  odd.  We 
may  avoid  this  awkward  distinction  by  writing  in  place  of  (3) 

(4)  ^.180°  +  C-1)"^6>, 

since  (—  1)"*  is  equal  to  +  1  or  —  1  according  as  m  is  even 
or  odd. 

Thus,  if  6  is  one  angle  whose  sine  is  equal  to  a  given  number 
«,  the  most  general  angle  which  has  the  same  sine  is 

^.180°+ (-1)"*  (9, 

where  m  is  any  positive  or  negative  integer  or  zero, 

A  brief  way  of  indicating  this  fact  is  given  by  the  follow- 
ing equations: 

(5)  sin  [m  •  180°  +  (_  1)^  6^]  =  sin  6 
or 

sin  [mir  +  (  —  1)""  ^]  =  sin  6, 

where  we  use  the  first  or  the  second  form  according  as  6  is 
expressed  in  degrees  or  in  radians.  Since  the  cosecant  of  an 
angle  is  the  reciprocal  of  its  sine,  we  have  also 

(6)  CSC  [m  .  180°  -h  (  -  1)"^  0^  =  esc  6 
or 

CSC  [WTT  +  (  —  1)"*  ^]  =  CSC  0. 

We  find,  by  precisely  similar  considerations, 

cos  (2  m  .  180°  ±6}=  cos  (9,  cos  (2  wtt  ±  (9)  =  cos  0, 

(^^  sec  (2  m  -  180°  ±6)=  sec  (9,  ^^  sec  (2  mir  ±6)=  sec  0, 
and 

tan  (m  •  180°  +  6)  =  tan  6,  ^     tan  (mir  +  6)  =  tan  B, 

^^)  cot  (m  .  180°  +  ^)  =  cot  6,  ^^     cot  (rmr  +  6>)  =  cot  6, 

according  as  the  angles  are  expressed  in  degrees  or  in  radians. 


INVERSE   TRIGONOMETRIC   FUNCTIONS  217 

EXERCISE    Llll 

Without  making  use  of  the  trigonometric  tables,  find  all  of  the  angles 
which  satisfy  the  following  conditions: 

1.    sin^=f  3.    tan^=l.  5.    cot  ^  =  VS.        7.   tan  ^  =  oo. 

/2.   cos^  =  -|.        4.   sec^  =  V2.        6.'^csc^=l.  8.  ^sin  ^  =  0. 

Making  use  of  the  tables,  find  all  of  the  angles  which  satisfy  the  fol- 
lowing equations : 

9.   sin  ^  =  -  .4721.  11.   tan  6  =  1.7269. 

10.   cos  ^  =  +  .3216.  12.   sec  ^  =  2.7213. 

97.   The  inverse  trigonometric  functions.     In  the  equation 

(1)  X  =  sin  y, 

we  now  propose  to  regard  y,  the  angle  or  arc,  as  a  function 
of  27,  the  sine.  We  may  express  this  new  way  of  looking 
at  the  relation  (1),  by  saying  that 

y  is  an  angle  whose  sine  is  equal  to  x 
or 

(2)  y  is  an  arc  whose  sine  is  equal  to  05, 

a  statement  which  is  usually  written  in  the  contracted  form 

(3)  y  —  arc  sin  oc. 

It  should  be  noted  that  (1)  and  (3)  are  merely  different 
ways  of  expressing  the  same  relation  between  x  and  y. 
They  differ  only  in  one  respect.  In  (V)  y  is  regarded  as 
given  and  x  is  to  be  found,  while  in  (8)  x  is  regarded  as 
given  and  y  is  to  be  found.  The  relation  between  the  func- 
tions (1)  and  (3),  that  of  being  inverses  of  each  other,  is 
of  the  same  kind  as  in  the  more  familiar  case  of  the  function 
X  =  y'^^  which  has  as  its  inverse  y  =  ±  ^x. 

In  the  same  way  we  define  the  equation 

y  =  arc  cos  a? 

to  mean  that  y  is  an  arc  whose  cosine  is  equal  to  x.  There- 
fore this  equation  is  equivalent  to 

X  =  cos  y. 


218  INVERSE   TRIGONOMETRIC   FUNCTIONS 

Similarly,  if  a;  =  tan  y^  we  write 

y  =  arc  tan  x, 
and  in  the  same  way  we  define  the  symbols 

arc  cot  sc,  arc  sec  oc,  arc  esc  oc. 

Let  us  return  to  equations  (1)  and  (3).  We  know  that 
the  sine  of  an  angle  is  never  numerically  greater  than  1. 
Consequently  we  can  find  no  angle  whose  sine  is  equal  to 
a;  if  a;  is  numerically  greater  than  1.  We  may  express,  this 
as  follows: 

77ie  function  arc  sin  x  is  defined  only  for  those  values  of  x 
which  are  not  numerically  greater  than  1 . 

If  X  is  numerically  less  than  unity,  there  exists  not  merely 
one  angle  whose  sine  is  equal  to  a?,  but  the  number  of  such 
angles  is  unlimited.  (See  Art.  96.)  If  x  is  positive,  one  of 
the  corresponding  angles  is  a  positive  acute  angle.  If  x  is 
negative,  the  smallest  corresponding  'positive  angle  is  in  the 
third  quadrant.  But  in  this  case  there  is  a  negative  acute 
angle  whose  sine  is  equal  to  the  given  negative  value  of  x. 

Let  us  use  the  symbol 

Arc  sin  nc, 

distinguished  from  arc  sin  x  hy  the  use  of  the  capital  letter  A, 

to  indicate  the  numerically  smallest  angle  or  arc  whose  sine  is 

equal  to  x. 

The  function  Arc  sin  x,  like  arc  sin  x,  is  defined  only  for 

the  values  of  x  between  —  1  and  +  1 ;  that  is,  for  those  values 

of  X  for  which  ^  ^        . 

-l<x<^l. 

But  for  every  such  value  of  x^  Arc  sin  x  has  only  one  value, 
while  arc  sin  x  has  infinitely  many  values.  For  positive 
values  of  x,  Arc  sin  a;  is  a  positive  acute  angle,  and  for  nega- 
tive values  of  a:  it  is  a  negative  acute  angle.     No  value  of  Arc 

sin  X  ever  exceeds  the  limits  ±  90°  or  ±  -  radians,  so  that 
we  shall  always  have 

(5)  ""  o"  —  -^^^  ^^^  ^  —  o"' 

2  Z 


I 


INVERSE   TRIGONOMETRIC   FUNCTIONS 


'219 


We  shall  henceforth  speak  of  Arc  sin  x  as  the  principal  value 
of  arc  sin  x^  and  we  know  from  equation  (5)  of  Art.  96  that 

(6)  arc  sin  x  =  mir  +  (—  1)"*  Arc  sin  x^ 
where  m  is  any  positive  or  negative  integer  or  zero. 

In  many  books  the  symbol  arc  sin  x  is  used  in  the  restricted 
sense  which  we  have  given  to  Arc  sin  x.  For  some  purposes 
the  distinction  between  the  two  functions  arc  sin  x  and  Arc 
sin  X  is  not  important.  But  for  certain  other  questions,  a 
careful  discussion  of  the  principal  value  is  the  only  way  to 
avoid  hopeless  confusion. 

This  whole  matter  will  become  very  clear  if  we  make  a 
graph  of  the  function 

(3)  y  =  arc  sin  x. 

Since  this  equation  between  x  and  y  has  the  same  signifi- 
cance as 

(1)  a;  =  sin  «/, 

we  may  plot  the  latter  relation  instead  of  (3).     But  we  have 
already  studied  the  graph  of  the  similar  equation 

(7)  y  =  sin  X  (see  Arts.  73  and  78), 

and  clearly  the  graph  of  (1)  may  be  obtained  from  that  of 
(7)  by  interchanging  x  and  t/.  In  other  words  the  graph  of 
(1),  or  what  amounts  to  the  same  thing,  the 
graph  of  (3),  is  a  sine  curve  placed  in  a  ver- 
tical position.     (See  Fig.  125.) 

The  graph  shows  clearly  that  the  function 


y  =  arc  sm  x 

is  not  defined  for  values  of  x  which  are  nu- 
merically greater  than  unity.  It  also  shows 
that  for  every  admissible  value  of  x  there  are 
an  infinite  number  of  values  of  y,  viz.,  the 
ordinates  of  all  of  the  points  Pj,  Pg^  -^s'  etc., 
in  which  a  line  parallel  to  the  «/-axis,  at  a  distance  a;,  inter- 
sects the  curve. 


*y 


.^. 


/ 


Fig.  125 


220  INVERSE   TRIGONOMETRIC   FUNCTIONS 

But  the  curve  which  corresponds  to  the  principal  value  of 
arc  sin  x^ 

'  y  =  Arc  sin  x^ 

consists  merely  of  that  portion  of  the  graph  of  arc  sin  x  which 
lies  between  the  points  A  and  B.  This  part  of  the  curve  is 
indicated  in  Fig.  125  by  a  heavier  line.* 

In  a  similar  way  we  define  a  principal  value  Arc  cos  x  for 
the  function  arc  cos  x.  The  function  arc  cos  x  is  defined  for 
all  values  of  x  which  are  not  numerically  greater  than  1,  and 
has  infinitely  many  values  for  every  admissible  value  of  x.  It  is 
convenient  to  define  as  its  principal  value  Arc  cos  a;,  the  smallest 
positive  angle  whose  cosine  is  equal  to  a:,  so  that  Arc  cos  x  is 
subject  to  the  following  inequalities 

0  ^  Arc  cos  aj  <  TT. 

The  detailed  discussion  of  these  statements  is  left  to  the 
student  as  an  exercise. 

It  should  be  remarked  that  the  notations  sin-*  a?,  cos-^a?, 
etc.,  are  also  in  use  for  arc  sin  a;,  arc  cos  x^  etc.  This  second 
notation,  which  is  frequently  used  in  other  branches  of 
mathematics,  has  the  advantage  of  emphasizing  the  fact 
that  sin  x  and  sin"^  x  are  inverse  functions  of  each  other. 
But  it  has  the  disadvantage  of  colliding  with  the  customary 
notation  for  exponents,  and  therefore  tends  to  create  con- 
fusion. Thus  we  usually  write  sin^rr  for  (sin  a;)^,  and  a~^ 
for  1/a.  Thus  the  symbol  sin'^  x  might  consistently  be  in- 
terpreted to  mean 

1 

-: =  CSC  X^ 

Sin  X 
which  is  something  entirely  different  from  arc  sin  x. 

*  If  we  had  chosen  as  principal  value  of  arc  sin  x  the  smallest  positive  angle 
whose  sine  is  equal  to  x,  the  principal  value  would  not  be  represented  by  a  con- 
tinuous (unbroken)  curve.  One  part  of  this  curve  would  be  OB,  and  the  other 
part  (corresponding  to  negative  values  of  x)  would  be  between  y  =  Trand  ?/  =  3  7r/2. 


INVERSE   TRIGONOMETRIC   FUNCTIONS  221 

EXERCISE  LIV 

1.  Show  that  the  function  y  =  arc  tan  x  is  defined  for  all  values  of 
X.  Draw  the  graph  of  the  function  and  show  that  its  principal  value 
may  be  selected  subject  to  the  conditions 

0  ^  Arc  tan  x<.ir. 

2.  Investigate  the  function  arc  cot  x  in  the  same  way. 

y 

3.  Are  the  functions  arc  sec  x  and  arc  esc  x  defined  for  all  values  of 
X  ?     Draw  their  graphs  and  choose  principal  values  for  these  functions. 

Find  the  values  of  the  following  expressions : 

4.  Arc  sin  \.  6.   Arc  tan  1,  arc  tan  1. 

5.  Arc  cos  \.  7.   Arc  cos  (^  V3),  arc  cos  (W^)- 

By  using  the  table  of  natural  functions  compute  the  values  of  the 
following  quantities : 

8.  Arc  tan  1.3722  +  Arc  cos  0.4321. 

9.  Arc  sin  0.3425  +  Arc  cot  1.7264. 

10.  Obtain  the  value  of  sin  (Arc  sin  \  +  Arc  sin  4). 

11.  Obtain  the  value  of  sin  (Arc  sin  f  +  Arc  cos  ^). 

12.  If  X  and  y  are  positive  numbers,  less  than  unity,  show  that 


sin  (Arc  sin  x  +  Arc  sin  ?/)  =  xVl  —  y^  +  ^ Vl  —  x^. 
Solution.     For  abbreviation  put 

Arc  sin  x  =  u,    Arc  sin  y  =  v.  y-'-"^ 

Since  x  and  y  are  positive  numbers,  less  than  unity,  w^d  v  will  be 
positive  acute  angles  such  that 
(1)  sin  u  =  x^  sin  v  =  y, 

and  therefore 


(2)  cos  u  =  Vl  —  x^,        cos  V  =  VI  —y^. 

We  now  find 

sin  (Arc  sin  x  +  Arc  sin  y)  =  sin  (u  +  v)=  sin  u  cos  v  +  cos  u  sin  v, 
and  therefore,  on  account  of  (1)  and  (2), 

sin  (Arc  sin  x  +  Arc  sin  y)  =  xVl  —  y^  +  y  Vl  —  x^. 


13.   Show  that  cos  (Arc   sin  x  —  Arc  sin  y)  =  Vl  —  x^  Vl  —  y^  +  xy 
if  O^xgl,  O^y^l. 


t 


14.    Show  that  Arc  sin  x  +  Arc  cos  ^  =^  if  — l<a:^  +  l. 


222  INVERSE   TRIGONOMETRIC   FUNCTIONS 


15.   Prove  the  formula  Arc  tan  x  —  Arc  tan  y  =  Arc  tan 


X  —  y 


l  +  xy 

16.  Do  the  equations  of  Exs.  12  and  13  undergo  any  modifications 
if  negative  values  of  x  and  y  are  admitted  ?  Discuss  in  order  the  cases 
a:<0,2/>0;  a:>0,y<0;  a;<0,2/<0. 

98.  Trigonometric  equations.  It  often  happens  that  angles 
are  to  be  determined  by  means  of  equations  which  they  must 
satisfy.  Such  equations  usually  contain  the  trigonometric 
functions  of  the  unknown  angle  and  are  then  known  as 
trigonometric  equations. 

Let  us  confine  our  attention  to  the  case  where  one  unknown 
angle  is  to  be  found  as  a  solution  of  one  equation.  Such  an 
equation  may  have  one  of  the  following  forms. 

I.  It  contains  6  algebraically,  but  does  not  contain  the 
trigonometric  functions  of  0. 

Example.  e^-~  =  0. 

II.  It  contains  the  trigonometric  functions  of  the  angle  d 
in  algebraic  combinations,  but  does  not  contain  the  angle  0 
itself  explicitly. 

Examples,     sin^  ^  —  cos  ^  =  0,   2  tan  ^  +  cot  ^  =  —  3. 

III.  It  contains  the  angle  0  and  its  trigonometric  func- 
tions simultaneously. 

Example.  ^  —  -  sin  ^  =  ^ . 


equ 
^se 


Clearly  the  equations  of  the  first  type  are  merely  algebraic 
equations,  and  their  discussion  properly  belongs  to  a  trea- 

se  on  algebra, 
t  may  be  shown  that  the  equations  of  the  second  type  can 
also  be  reduced  to  algebraic  equations,  although  it  is  often 
easier  to  effect  their  solution  without  so  reducing  them. 

The  equations  of  the  third  type  will  not  be  considered  in 
this  book.  The  solution  of  such  equations  is  a  difficult 
matter  and  can  be  accomplished  in  a  satisfactory  manner 
only  in  a  few  cases.     It  should  be  mentioned,  however,  that 


TRIGONOMETRIC   EQUATIONS  223 

the  method  of  graphs  usually  provides  an  approximate  solu- 
tion for  such  equations. 

We  shall  now  discuss  a  few  simple  examples  of  equations 
of  the  second  type,  in  such  a  way  as  to  illustrate  the  fact 
that  their  solution  may  be  reduced  to  that  of  algebraic 
equations. 

EXERCISE    LV 

1.  Solve  the  equation  2  cos^  x  —  5  +  7  sin  x  =  0. 

Solution.  We  have  cos^  x  =  1  —  sin^  x.  Therefore  the  given  equation 
becomes  -  3  -  2  sin2  x  +  7  sin  a;  =  0. 

Consequently,  if  we  put  sin  x  =  s,  we  obtain  the  quadratic  equation 
iors,  2  s2  _  7  s  +  3  =  0. 

The  solution  of  this  equation  gives 

s  =  3  or  s  =  |. 
The  first  solution  must  be  discarded,  since  s  =  sin  x  cannot  be  numerically 
greater  than  unity.     The  second  solution  tells  us  that 

sin  X  =  ^, 
so  that  X  =  30°  and  x  =  150°  are  the  only  positive  angles  smaller  than 
360°  which  satisfy  the  equation. 

In  examples  2  to  5,  find  all  positive  angles  less  than  360°  which  satisfy 
the  given  equations.     This  may  be  done  without  using  any  tables. 

2.  cos  2  ^  +  cos  ^  =  -  1.  4.   tan  2  ^  =  -  2  sin  0. 

3.  cot  2  ^  +  tan  ^  =  -  f  V3.  5.   sec2  $  +  csc^  0  =  4:. 
In  solving  the  following  equations,  make  use  of  the  tables : 
6.    sin  a:  tan  a;  ::=  —  ^^^.                             7.    cos  a;  cot  a:  =  —  f . 
8.    Solve  the  equation  cos  3  a:  =  sin  2  x. 

99.   The  equation  a  sin  8  +  6  cos  9  =  c.     The  equations  of 
the  form 
(1)  a  sin  0  -\-b  cos  6  =  e 

may  be  solved  by  the  method  of  the  preceding  article.  But, 
unless  the  numbers  a,  6,  c  are  especially  simple,  it  is  far  more 
convenient  to  proceed  as  follows : 

We  introduce  two  auxiliary  quantities  I  and  X,  such  that 

.ry.  I  sin  L  =  a, 

I  cos  L  =  b, 


224 


INVERSE  TRIGONOMETRIC   FUNCTIONS 


where,  moreover,  I  is  assumed  to  be  positive.     That  it  is 
always  possible  to  find  a  positive  number  I  and  an  angle  L 

which  satisfy  equations  (2)  be- 
comes obvious  if  we  think  of  a 
and  h  as  the  rectangular  coordi- 
nates of  a  point  P.  (See  Fig. 
126.)  Equations  (2)  show  that  I 
is  the  radius  vector  OP  of  P  and 
that  L  is  the  angle  which  OP 
makes  with  the  positive  direction 
of  the  ic-axis.  Our  figure  also 
shows  us  that 


+a? 


Fig.  126 


(3) 


tanX  =  ^,    l  =  +^a^  +  h\ 

0 


Of  course  these  same  equations  also  follow  directly  from  (2) 
without  the  use  of  geometry. 

If  now  we  substitute  the  values  (2)  for  a  and  h  in  (1),  we 

^^^  I  (sin  i  sin  ^  +  cos  L  cos  6)  =  c, 

or  by  Art.  79,  equation  (8), 

(4)  Zcos(l9-X)=(?. 

Therefore,  in  order  to  solve  (1)  we  may  first  determine  I 
and  L  from  (2)  and  then  find  6  from  (4). 


EXERCISE    LVI 
1.   Solve  the  equation  2.1346  sin  6  -  3.0526  cos  0  =  0.9875. 


Solution. 


a  =  +  2.1346 
h  =  -  3.0526 

c  =  +  0.9875 


log  a  =  log  (I  siu  L)  =  0.32932 

(4) 

log  b  =  log  (I  cos  L)  =  0.48467  n 

(5) 

log  tan  Z  =  0.81465  n 

(7)  =  (4) -(5) 

L  =  145°  2M5 

(8) 

log  sin  i  =  9.75820 

(9) 

log  cos  Z  =  9.91355  n 

(10) 

log/  =  0..57112 

(11)  =  (4) -(9)  =  (5) -(10) 

logc  =  9.99454 

i^) 

log  cos  (d-L)=  9.42342 


(1) 
(2) 
(3) 

=  ( 

) 
)  = 

(12)  =  (6) -(11) 


TRIGONOMETRIC   EQUATIONS  225 

e^-  L=    74°  37'.61    (13) 

^2 -i  =  285°  22'. 39   (14) 
L  =  145°    2M5   (8) 
^1  =  219°  39'.76   (15)  =  (13) +  (8) 
$2=    70°24'.54  (16)  =  (14) +  (8) 

Remarks.  The  numbers  in  parentheses  indicate  the  order  in  which 
the  results  are  written  down  and  how  some  of  them  are  obtained.  The 
—  10  has  not  been  written  down  in  the  case  of  negative  characteristics. 
The  n  which  follows  the  logarithms  of  b,  tan  L  and  cos  L  indicates  that 
the  corresponding  numbers  are  negative.  (See  Art.  25.)  L  is  chosen  in 
the  second  quadrant  because,  I  being  positive,  sin  L  is  positive  and  cos  L 
is  negative.  The  values  6^  —  L  and  6^  —  L,  both  obtained  from  (12), 
are  the  two  values,  less  than  360°,  which  0  —  L  may  have  so  as  to  corre- 
spond to  the  given  value  (12)  of  log  co8(^  —  Z), 
Check.     By  substitution  in  the  original  equation, 

log  a  =0.32932  log  &  =  0.48467  n 

log  sin  ^1        =  9.80500  n  log  cos  6^        =  9.88639  n 

log  (a  sin  6^)  =  0713432  n  log  {h  cos  ^i)  =  0.37106 

a  sin  ^1  =  -  1.3624 

h  cos  e^  =  +  2.3500 

a  sin  e^  +  h  cos  $]_  =  +  0.9876 

c  =  4-  0.9875 

This  calculation  checks  6i  and  therefore  also  L.     The  relation  between 
^1,  62-,  and  L  is  so  simple  as  to  make  unnecessary  a  separate  check  for  9^. 

2.  Solve  the  equation 

-  3.2471  sin  6  +  5.7469  cos  ^  =  -  6.3271. 

3.  Solve  the  equation 

2.1725  sin  6  +  3.2749  cos  $  =  5.7216. 


CHAPTER   XIV 


APPLICATIONS   TO   THE   THEORY   OF   WAVE  MOTION 

100.  Simple  harmonic  motion.  When  we  began  to  study 
trigonometry,  it  was  for  a  very  practical  purpose.  We 
wished  to  find  an  arithmetical  method  for  solving  triangles. 
We  accomplished  this  purpose  by  means  of  the  trigonometric 
functions  and  by  using  tables  of  the  numerical  values  of  these 
functions.  Later  we  generalized  the  notion  of  the  trigono- 
metric functions  more  than  was  strictly  necessary  for  the 
simple  problem  of  solving  triangles,  and  we  found  it  to  be  an 
interesting  task  to  investigate  these  trigonometric  functions 
and  their  various  properties  for  their  own  sake.  We  shall 
now  find  that  these  properties,  aside  from  their  theoretical 
interest,  have  in  their  turn  most  important  applications. 

The  fundamental  reason  for  the  great  importance  of  the 
trigonometric  functions  lies  in  their  periodicity.  (Cf. .Art. 
67.)  Many  natural  phenomena  are  periodic  in  character, 
and  whenever  the  attempt  is  made  to  represent  such  a  phe- 
nomenon by  a  mathematical  expression, 
the  trigonometric  functions  are  found 
to  be  indispensable. 

The  simplest  periodically  recurring 

motions  are  connected  with  uniformly 

\a~'^^  rotating    bodies.      Let    the    point 

(Fig.  127)  describe  a  circular  path  o\ 
radius  a  around  the  point  0  as  center| 
and  let  us  assume  that  it  is  moving 
with  a  constant  angular  velocity  of 
radians  per  second.  Let  us  assume 
further  that  P  starts  its  motion  at  the  time  ^  =  0  from  th( 
point  (7,  which  is  so  located  that  Z  ^  0(7  is  equal  to  a  radians. 

226 


SIMPLE   HARMONIC  MOTION  227 

If  P  is  the  position  of  the  point  at  the  time  <,  that  is,  t 
seconds  after  the  motion  has  begun,  we  shall  have 

(1)  e  =  a}t+a, 

where  6  denotes  the  angle  A  OP  expressed  in  radians.  The 
point  will  describe  its  circular  path  in  counterclockwise  or 
clockwise  fashion  according  as  ©  is  positive  or  negative. 

While  the  point  P  is  moving  in  its  circular  path,  its  pro- 
jection M  upon  the  a;-axis  will  oscillate  to  and  fro  between 
the  points  A  and  A\  reaching  its  greatest  speed  when  at  0, 
gradually  slowing  down  until  it  reaches  A\  when  it  reverses 
the  direction  of  its  motion,  returns  with  gradually  increasing 
speed  to  0,  after  reaching  which  point  it  slows  down  again 
until  it  reaches  A  and  again  reverses  its  motion. 

To  find  an  analytic  expression  for  the  motion  of  the  point 
M^  we  observe  that 

OM^x,     OP=a,     A0P  =  6,     0M=  OP  cos  AOP, 

whence,  making  use  of  (1), 

(2)  a:=a  cos  (a)^  + a). 

In  the  same  way  we  see  that  iV,  the  projection  of  P  upon 
the  ^-axis,  moves  in  accordance  with  the  equation 

(3)  ^  =  a  sin  (ft)^+ a). 

Equations  (2)  and  (3)  are  so  closely  related  that  it  will 
suffice  to  study  one  of  them.     In  fact,  if  we  put  in   (3) 

a  =  ^  +  a^  it  becomes  (cf.  Art.  77,  equations  (2)), 
y  =  a  sin  la)t-\-a^-{-  —  j  =  a  cos  (^cot  +  a'), 

which  is  of  the  same  form  as  (2).  We  shall  therefore  con- 
fine our  attention  to  equation  (3). 

Since  any  diameter  of  the  circle  may  be  chosen  as  y-axis, 
we  may  express  our  result  as  follows.  If  a  point  describes  a 
circular  path  with  uniform  velocity/,  its  projection  upon  any 
fixed  diameter  of  the  circle  moves  in  accordance  with  an  equa- 
tion of  the  form  (3).  Such  a  motion  is  called  a  simple  har- 
monic motion.     The  quantity  a  which  measures  the  maximum 


228  THEORY  OF   WAVE  MOTION 

distance  of  the  point  iV  from  its  mean  position  0  is  called  the 
amplitude. 

101.  The  period  and  phase  constant.  When  the  angle  B 
has  increased  from  its  initial  value  a  by  2  tt  radians,  the 
point  P  will  have  described  a  complete  circumference  and 
the  motion  of  the  point  N  will  have  passed  through  all  of 
its  phases.  The  time  T^  which  is  required  to  accomplish 
this,  is  called  the  period  of  the  simple  harmonic  motion.  The 
period  is  determined  by  the  condition  that  the  angle  (oT  de- 
scribed by  the  point  P  in  the  time  T  must  be  equal  to  2  tt. 
Therefore  we  find 

(1)  G)^=2  7r,   T=^. 

If  we  wish  to  put  the  period  into  evidence  in  the  equation 
of  a  simple  harmonic  motion,  we  observe  that  (1)  gives 

27r 
T 

so  that  we  may  write,  in  place  of  (3),  Art.  100, 

(2)  ^  =  «sin^-^  +  aJ. 

This  equation  represents  a  simple  harmonic  motion  of 
period  T  and  of  amplitude  a.  The  quantity  a  is  called  the 
phase  constant.  The  phase  constant  is  an  angle  and  enables 
us  to  calculate  the  distance  from  0  to  the  position  occupied  by 
the  moving  point  iVat  the  time  ^  =  0  when  the  motion  began. 

Thus,  if  a  =  0,  the  point  N  starts  from  0  as  its  initial  position  and 
begins  to  move  upward ;  if  «  =  -,  the  initial  position  of  N  is  at  B,  etc. 

Let  us  think  of  two  different  points  oscillating  up  and 
down  along  the  line  BB\  each  in  accordance  with  an  equation 
of  the  form  (2),  the  amplitudes  and  periods  of  the  two  mo- 
tions being  the  same  while  the  phase  constants  are  different. 
Then  these  two  points  will  move  in  exactly  the  same  way, 
only  that  one  will  always  be  ahead  of  the  other.  The  sec- 
ond point  will  appear  to  be  imitating  the  motion  of  the  first, 


ILLUSTRATIONS   OF   HARMONIC  MOTION  229 

lagging  behind  it  in  a  perfectly  definite  fashion.  This  is 
what  is  meant  by  saying  that  the  phases  of  the  two  motions 
are  different,  the  phase  of  the  motion  (2)  at  the  time  t  being 
equal  to  ^    ^ 


T 


+  a. 


The  use  of  the  word  phase  in  this  connection  is  not  merely  accidental. 
The  appearance  of  the  moon  at  a  given  instant  (its  phase)  depends 
upon  the  place  in  its  orbit  around  the  earth  which  it  happens  to  occupy 
at  that  time.  By  analogy  we  speak  of  a  periodic  phenomenon  as  passing 
through  all  of  its  phases  in  the  course  of  a  period,  and  of  course  two 
otherwise  identical  periodic  phenomena  may  have  their  corresponding 
phases  occur  at  different  times.  It  is  this  difference  which  manifests 
itself  in  the  different  values  of  the  phase  constant. 

We  have  already  observed  that  the  substitution  «  =  —  +«' 

converts  (3),  Art.  100,  into  an  equation  of  form  (2),  Art.  100. 
We  may  now  express  this  fact  as  follows: 

The  two  equations 


Vi 


=  a  sm  -^,  ^2  =  «  cos  -^=  a  sin  (^^r+  2J 


represent  two  simple  harmonic  motions  of  amplitude  a  and 
period  T^  which  differ  only  in  phase^  the  phase  difference  being 

equal  to  —  radians  or  90°. 

The  time  interval  which  elapses  between  corresponding 
phases  of  these  two  motions  is  \  T^  that  is,  a  quarter  period. 

102.   Some  illustrations  of   simple  harmonic  motion.     The 

notion  of  simple  harmonic  motion  is  of  fundamental  impor- 
tance in  many  problems  of  applied  mathematics.  The  mo- 
tion of  a  simple  pendulum,  the  vibrations  of  a  tuning  fork, 
and  many  of  the  motions  of  elastic  bodies  may  be  described 
conveniently  in  terms  of  simple  harmonic  motion.  The 
vertical  motion  of  a  particle  of  a  water  wave  is  approximately 
of  the  same  type,  and  the  whole  theory  of  sound  and  light  is 
based  on  the  idea  of  harmonic  motion. 


230  THEORY   OF   WAVE  MOTION 

EXERCISE  LVII 

In  Exs.  1-5  the  unit  of  time  is  one  second  and  the  nnit  of  length 
one  inch.  Describe  completely  the  simple  harmonic  motion  given  by 
each  of  these  equations;  that  is,  determine  their  amplitudes,  periods, 
and  phase  constants,  assuming  equation  (2)  of  Art.  101  as  the  standard 
form  for  the  equation  of  such  a  motion. 

1.  2/ =  2.3745  sin  ?-^. 
^  5 

2.  y  =  3.7216  sin  (4  <  +  ^y 

3.  y  =  1.4712  sin  (2.7215 1  +  1.7291). 

4.  y=  11.7261  cos  (7  0- 


5. 


6.  A  certain  pendulum  has  a  period  of  oscillation  of  5  seconds.  Its 
motion  is  started  by  displacing  the  bob  from  its  position  of  equilibrium 
three  inches  toward  the  right  and  then  releasing  it.  Write  the  equation 
of  its  motion.  What  will  be  the  corresponding  equation  for  a  point 
halfway  between  the  lower  end  of  the  pendulum  and  its  point  of 
support  ? 

Hint  Since  the  amplitude  of  the  oscillation  is  small  as  compared 
with  the  length  of  the  pendulum,  the  motion  may  be  regarded  as  taking 
place  approximately  in  a  straight  line.  Take  this  line  as  z-axis,  positive 
toward  the  right,  and  choose  as  origin  the  point  of  equilibrium.  Let  the 
time  t  be  measured  in  seconds  from  the  moment  in  which  the  pendulum 
is  released.  Then  the  required  equation  for  the  lower  end  of  the  pendu- 
lum will  be 

7.  A  cork  is  bobbing  up  and  down  owing  to  the  passing  water  waves. 
These  waves  are  4  inches  high  (i.e.  the  difference  of  level  between  the 
crest  and  trough  is  4  inches).  If  seventy  of  these  waves  pass  in  a 
minute,  and  we  start  to  count  time  from  one  of  the  instants  when  the 
cork  has  reached  its  highest  position,  what  equation  will  describe  the 
motion  of  the  cork  approximately  ? 

8.  Show  that  the  following  mechanism  enables  us  to  convert  uniform 
circular  motion  into  simple  harmonic  motion.  RR'  (Fig.  128)  is  a  rod 
which  may  slide  back  and  forth  in  its  own  direction,  its  motion  being 
limited  by  the  guides  A,  B,A',  B'.  Attached  to  the  rod  there  is  a  slot  S 
perpendicular  to  the  rod.   A  crank  C,  moving  with  uniform  velocity  in  a 


SIMPLE   HARMONIC   CURVES 


231 


circle  around  the  point  O,  is  made  to  fit  accurately  into  the  slot  S,  Every 
point  of  the  rod  RR'  will  then  describe  a  simple  harmonic  motion. 


c 


B' 


9 


Fig.  128 

9.   Show  that  a  point  on  the  piston  rod  of  a  steam  engine  will  move  to 
and  fro  approximately  according  to  the  law  of  simple  harmonic  motion. 

103.   Simple   harmonic  curves.     In  the    equation  charac- 
teristic of  a  simple  harmonic  motion,  namely, 

y  =  a  sin  (o)^  -f  «), 

let  us  substitute  x  in  place  of  t  and  interpret  x  and  y  as  the 
rectangular  coordinates  of  a  point  in  a  plane.  The  curves 
obtained  as  a  result  of  plotting  such  equations, 

^=  a  sin(o)a;  +  a), 

are  called  simple  harmonic  curves. 

We  may  also  think  of  the  relation  between  simple  harmonic  motion 
and  simple  harmonic  curves  in  the  following  more  concrete  fashion.  At- 
tach a  light  pin  P  (Fig.  129)  to  one  of  the 
prongs  of  a  vibrating  tuning  fork,  and  allow 
it  to  press  lightly  against  a  strip  of  smoked 
glass.  If  this  strip  of  glass  is  at  rest,  the  pin 
will  make  a  short  straight  line  upon  it.  But 
if  the  strip  be  moved  with  a  constant  velocity 
in  a  direction  perpendicular  to  that  of  the 
vibration   of   the  tuning  fork,  the  point  P 

will  describe  a  wavelike  curve  upon  the  slide.  This  curve  may  be 
regarded  as  a  record  of  the  motion  of  the  tuning  fork.  It  is  easy  to  see 
that  the  record  produced  in  this  way  by  any  simple  harmonic  motion  is  a 
simple  harmonic  curve. 

The  simplest  case  of  a  simple  harmonic  curve  is  that  of 
the  sine  curve  ^  =  sin  a:. 


Fig.  129 


232  THEORY  OF   WAVE   MOTION 

whose  form  has,  by  this  time,  become  familiar  to  the  student. 
(Compare  the  middle  curve  in  Fig.  130.)  It  has  the  form 
of  a  wave  line  with  nodes  at  the  points  a;  =  0,  a:  =  ±  tt, 
x=  ±  2  TT,  etc.,  with  crests  or  maxima  one  unit  high  above  the 

of  the  ic-axis,  and  with  troughs  or  minima  one  unit  deep 
below  the  points 

3  TT  7  TT  11  TT        , 

a:  =  — ,  ^  =  "2-'  ^=  -2"'  ®^^-' 

The  length  of  one  complete  undulation  of  the 
curve  is  equal  to  2  tt. 

104.  Amplitude.     It  is  clear 
that  the  curve 

(1)  y  =  (^  sin  27, 

where  a  is  any  fixed  positive 
number,  is  of  the  same  general 
form  as  the  sine  curve.  It 
has  the  same  nodes  (points  of 
intersection  with  the  a:-axis), 
and  each  of  its  undulations  has  the  same  length  as  that  of 
the  sine  curve.  Its  maxima  are  above  the  same  points  of  the 
rc-axis  as  those  of  the  sine  curve,  but  they  are  higher  or  lower 
according  as  a  is  greater  or  less  than  unity,  a  is  called  the 
amplitude  of  the  curve.  Figure  130  shows  three  such  curves 
of  amplitude,  ^,  1,  and  2.  It  is  clear,  then,  that  the  depth  of 
the  wavelike  curve  (1)  is  dependent  upon  the  value  of  the 
amplitude. 

105.   Wave  length.    The  two  curves 

y  =  sin  X  and  y  =  sin  2  a; 
are  shown  together  in  Fig.  131.     The  latter  curve  has  the 
same  general  form  as  the  former,  but  each  of  its  undulations 
(its  wave  length)  is  only  half  as  long. 

Similarl}^  the  curve 
(1)  i/  =  sinnx, 


PHASE  CONSTANT 


233 


where  ?^  is  a  positive  integer,  is  found  to  be  a  wave  line  of 
the  same  height  as  the  sine  curve,  but  having  n  complete 
undulations     between    x  =  0 
and  a;  =  2  TT.      Consequently 
its  wave  length  X  is  given  by 

n 


(2) 


Fig.  131 


Equation    (1)    represents    a 
simple    harmonic    curve     of 

2  TT 

wave-length  \  = — ,  even  if 
n 

n  is  not  an  integer.     For  the  function  sin  nx  will  pass  through 
all  of  its  values  jusH  once,  while  nx  changes  from  0  to  2  tt  ; 

2  TT 

that  is,  while  x  changes  from  0  to . 

n 

We  may  put  the  wave  length  into  evidence  in  the  equation 

of  the  curve,  by  solving  (2)  for  n  and  substituting  the  result- 


ing value  of  n  in  (1). 
(3) 


We  find  n  = ,  and  therefore 

X 


y  =  sm- 


TTX 


as  the  equation  of  a  simple  harmonic  curve  of  wave  length  X. 
If  we  combine  this  result  with  that  of  Art.  104,  we  see  that 


(4) 


y  =  a  sm 


represents  a  simple  harmonic  curve  of  wave  length  X  and  of 
amplitude  a, 

106.   Phase  constant.     If  we  compare  the  curves 

y  =  sin  X  and  y  =  cos  x, 

we  observe  that  they  are  identical  in  form.  That  is,  both 
are  simple  harmonic  curves  of  the  same  wave  length  and 
amplitude.  They  differ  only  in  position.  We  can,  in  fact, 
slide  one  of  these  curves  along  the  x-axis  in  such  a  way  as  to 


234  THEORY  OF   WAVE  MOTION 

make  it  coincide  with  the  other.  This,  as  we  have  observed 
before  (Art.  78),  is  the  geometrical  significance  of  the 
equation 

sin  l  —  -\-x]  =  cos X. 

We  may  therefore  dispense  with  the  cosine  curve  altogether 
and  regard  it  as  a  displaced  sine  curve.  More  generally,  the 
same  thing  is  true  of  the  curve 

(1)  ^  =  sin  (2:  + a), 

which  coincides  with  the  sine  curve  if  a  =  0  and  with  the 

cosine  curve  if  «  =  -5-.     Equation  (1)  represents  a  sine  curve 

displaced  toward  the  left  through  a  distance  of  a  units.  The 
quantity  a  is  called  the  phase  constant  of  this  curve. 

If  we  combine  the  results  of  Arts.  104  and  105  with  our 
latest  remark,  we  see  that 

(2)  y  =  a  sm  (^^—  +  « 1 

represents  a  simple  harmonic  curve  of  wave  length  X,  amplitude 

a,  and  phase  constant  a. 

This  equation  may  be  put  into  a  different  form  by  making 

use  of  the  addition  formula  of  the  sine.    (See  Art.  79.)     For 

we  have 

r  .     27rx             ,          2irx    .      ~l 
v=a\  sin cos  a  +  cos sm  a  . 

Consequently,  if  we  put 

(3)  m  =  a  cos  a,   w  =  a  sin  a, 

we  shall  find 

,..  .    2  Trrr  ,  2  7rx 

(4)  y  =  msm hwcos . 

Conversely^  any  equation  of  the  form  (4)  has  a  simple  har- 
monic curve  as  its  graph. 


WAVE  MOTION  235 

For  if  m  and  n  are  given,  we  may  compute  a  and  a  from 
(3),  thus  obtaining  the  value  of  amplitude  and  phase  con- 
stant.    Since  a  is  positive,  we  have,  from  (3),         ^ 

(5)  a=  +  Vm2  +  r^2,    tana  =  — 

the  quadrant  of  a  being  determined  from  the  sign  of  its  sine 
and  cosine  as  given  by  (3). 

EXERCISE    LVIII 

Draw  the  simple  harmonic  curves  which  correspond  to  the  equations 
given  in  Exs.  1  to  5 : 

1.  3^  =  2sin(3a;  +  ^y  3.   y  =  3.1  sin  (7.6  x  +  6.2). 

2.  y=3cos(2a;-|).  4.  y  =  2.8sin  (1^+72°). 

« 

5.  2/  =  sin  (a:  +  tt),  y  =  —  sin  x. 

6.  In  the  general  theorj'^  of  simple  harmonic  curves  we  assumed  a  to 
be  a  positive  quantity.  Show  that  this  assumption  does  not,  after  all, 
really  exclude  from  consideration  those  cases  in  which  a  is  negative.  In 
other  words  show  that  a  simple  harmonic  curve,  for  which  a  is  negative, 
coincides  with  another  one  for  which  a  is  positive  and  whose  phase  con- 
stant differs  from  that  of  the  first  curve  by  tt  radians. 

7.  Write  the  equations  of  the  curves  of  Exs.  1  to  5  in  the  form  (4)  of 
Art.  106. 

8.  A  simple  harmonic  curve  is  given  by  the  equation 

y  =  2.75  sin  ?^^  +  3.76  cos  ^^  • 
^  5.76  5.76 

Determine  its  wave  length,  amplitude,  and  phase  constant. 

9.  Discuss  in  the  same  way  the  equation 

y  =  3.72  cos  (7.52  x)-  2.67  sin  (7.52  x). 

107.  Wave  motion.  Our  use  of  the  word  wave^  in  con- 
nection with  simple  harmonic  curves,  is  not  quite  in  accord- 
ance with  the  accepted  meaning  of  this  term.  Ordinarily 
when  we  speak  of  a  wave,  a  water  wave,  for  example,  we 
mean  a  peculiar  kind  of  motion.  If  the  cross  section  of  the 
surface  of  the  water  at  a  given  instant  is  a  simple  harmonic 


236  THEORY  OF  WAVE  MOTION 

curve,  we  should  properly  speak  of  this  curve,  not  as  the 
wave,  but  as  the  instantaneous  profile  of  the  wave.*  It  is 
characteristic  of  a  wave  that  this  profile  is  in  motion. 

We  shall  obtain  an  excellent  idea  of  wave  motion  by 
allowing  a  simple  harmonic  curve  to  glide  along  the  a;-axis 
with  a  uniform  velocity  v.  We  shall  call  such  a  wave  a 
simple  harmonic  wave,  and  v  its  velocity  of  propagation. 

Let  t  denote  the  time  (expressed  in  seconds),  and  let  us 
assume  that  v,  the  velocity  of  propagation  (expressed  in  feet 
per  second),  is  positive,  so  that  the  wave  advances  in  the 
direction  of  the  positive  a>axis.  Let  the  simple  harmonic 
curve 

(1)  y  =  asm{—-  +  a] 

be  the  wave  profile  at  the  time  t.  The  profile  of  the  wave 
at  the  time  t  =  X)  (t  seconds  earlier)  was  a  curve  of  the  same 
form  as  (1),  but  situated  farther  toward  the  left.  Therefore 
its  equation  can  differ  from  (1)  only  in  the  value  of  the 
phase  constant.     Consequently  we  may  assume  that 

(2)  y  =  a  sin  1^—-  +  (^oj 

is  the  equation  of  the  wave  profile  at  the  time  ^  =  0.  We 
wish  tp  find  the  relation  between  a,  a^,  v,  X,  and  t. 

The  nodes  of  the  wave  profile  at  the  time  ^  =  0,  that  is,  its 
intersections  with  the  a:-axis,  are  obtained  from  (2)  by 
equating  y  to  zero.  These  nodes  (see  Fig.  132),  the  points 
Mq,  Mq,  Mq'\  etc.,  are  infinite  in  number,  and  the  distance 
between  two  consecutive  ones  is  equal  to  J  X  or  one  half  of 
the  wave  length.  One  of  these  nodes,  Mq  say,  will  be  ob- 
tained by  equating   — ^  +  «o  ^^  ^^^^*     Since  OMq  is  the 

A. 

abscissa  of  the  point  Mq  and  since  for  this  point 
— h  "o  =  ^' 


*  The  wave  profile  is  the  cross  section  which  one  would  obtain  of  the  surface 
of  the  water  if  it  were  to  freeze  suddenly  while  a  wave  is  passing. 


WAVE  MOTION 


237 


we  shall  have 
(3) 


Oitfo  =  -|^ 


IT 


After  t  seconds,  the  wave  profile  has  moved  from  its  original 


position  MqAqMq'  •  •  •  to  MAM 
profile  is  now  given  by  (1). 
The  nodes  of   this  profile 
are  the  points  of  the  curve 
for  which 


The  equation  of  the  wave 


2  7rrc 


+  a  =  JCTT, 


Fig.  132 


where  Jc  is  either  equal  to 

zero    or   to    a    positive   or 

negative  integer.     Consequently  these  nodes  are  the  points 

of  the  a;-axis  whose  abscissas  have  the  values 


x  = 


Xa 


IT 


+ 1  k\  where  Jc  =  0,  ±1,  ±2,  ±3, 


One  of  these  points  is  the  new  position  M  occupied  by  the 
point  Mq  as  a  consequence  of  the  motion  of  the  wave  profile 
from  MqAqMq^  ...  to  MAM'  •••.  Since  OM  is  the  abscissa  of 
M^  we  shall  therefore  have 


(4) 


OM^-^  +  ikX, 


where  A;  is  a  definite  positive  or  negative  integer  or  zero 
whose  precise  value  remains  to  be  determined. 

But  if  V  is  the  velocity  of  propagation  of  the  wave  in  feet 
per  second,  every  point  on  it  moves  through  a  distance  of  vt 
feet  in  t  seconds.     Therefore 


(5) 


OM-  OMq  =  vt. 


If  we  substitute  in  this  equation  the  values  (3)  and  (4)  foi 
OM  and  OMq,  ^ 

(6)  ii:ia_4n+ i^\  =  i,^. 


^       ^«    .    1  Z.X 
27r"2^+^~^^ 


238  THEORY  OF   WAVE  MOTION 

This  equation  must  be  true  for  all  values  of  t.  In  particular 
it  must  be  true  for  <=  0.  But  for  ^  =  0  we  have  a=  Wq,  so 
that  the  equation  involves  a  contradiction  unless  ^  =  0. 
Consequently  the  integer  h  which  appears  in  equations  (4) 
and  (6)  must  be  equal  to  zero,  and  we  have 


whence 
(7) 


2irvt 


If  we  substitute  this  value  of  a  in  (1),  we  find 
(8)  y  =  a^m\-^(x-vf)+aA 

as  the  general  equation  of  a  simple  harmonic  wave  of  amplitude 
a  and  wave  length  \,  whose  velocity  of  propagation  is  equal  to  v. 
We  may  still  speak  of  Uq  as  the  phase  constant.  It  is  the 
phase  constant  of  the  wave  profile  at  the  time  t  =  0.  The 
phase  constant  of  the  profile  curve  at  any  other  instant  may 
be  computed  from  (7). 

If  in  (8)  we  assign  a  fixed  value  to  t,  we  obtain  the  equa- 
tion of  the  wave  profile  at  that  instant.  Let  us  instead 
assign  a  fixed  value  to  rr,  so  that  y  becomes  a  function  of  t 
alone.  In  the  case  of  a  water  wave  this  would  amount  to 
a  study  of  the  upward  and  downward  oscillations  of  a  cork 
floating  upon  the  water.  We  shall  naturally  inquire  as  to 
the  length  of  time  which  is  required  to  complete  such  an 
oscillation.  This  time  is  called  the  period  and  may  be 
denoted  by  T.  Clearly  T  is  the  time  which  must  be  added 
to  t  so  as  to  change  the  argument  of  the  sine  function  in  (8) 
by  ±  2  TT.     Now  if  we  increase  t  hy  T  without  changing  x^ 

this  argument  changes  by ,  and  this  will  be  equal  to 


2  7rif 


vT  .  ^  \ 
—  =1,  or  r=  -. 
\  V 


GENERAL   HARMONIC  MOTION  239 

Therefore,  if  v  is  the  velocity  of  propagation^  \  the  wave 
lengthy  and  T  the  period^  we  have  the  relation 

(9)  X=Tv, 
Instead  of  (8)  we  may  now  write 

(10)  y  =  a8m[2^(|- !)  +  «„] 

as  the  general  equation  of  a  wave  of  length  \,  period  T^  ampli- 
tude a,  and  phase  constant  a^. 

We  see  finally  that  a  simple  harmonic  wave  has  for  its  profile 
at  any  moment  a  simple  harmonic  curve^  and  that  every  point 
upon  it  oscillates  up  and  down  in  accordance  with  the  law  of 
simple  harmonic  motion. 

EXERCISE    LIX 

If  the  units  of  length  and  time  are  feet  and  seconds  respectively,  com- 
pute the  wave  length,  period,  amplitude,  and  phase  constant  of  each  of  the 
waves  represented  by  the  following  equations  : 

1.   ,  =  3sin[2.(|-±)  +  |]. 
3.   y  =  a  sin  (bx  +  ct  -\-  d). 

108.  General  harmonic  motion.  A  simple  harmonic  motion 
is  the  simplest  case  of  an  oscillatory  phenomenon,  and  many 
natural  periodic  motions  may  be  adequately  described  as 
simple  harmonic  motions.  We  have  already  given  some  ex- 
amples of  such  cases.  But  very  frequently  the  motion, 
although  of  an  oscillatory  character,  is  more  complicated. 
In  the  case  of  a  water  wave,  for  instance,  we  see  that  the 
profile  of  the  wave  is  not  a  simple  harmonic  curve,  but  that 
there  are  smaller  waves  (ripples  so  to  speak)  running  along 
the  backs  of  the  larger  ones,  thus  complicating  the  motion. 
Simple  experiments  show  that  the  sound  waves  produced  by 
a  tuning  fork  are  very  approximately  represented  by  simple 
harmonic  motion;  but  other  musical  instruments,  such  as  the 


240  THEORY  OF  WAVE   MOTION 

violin,  the  piano,  the  human  voice,  produce  sound  waves 
which  resemble  the  more  complicated  water  waves. 

A  tuning  fork  which  makes  129  oscillations  in  a  second 
causes  a  certain  simple  harmonic  motion  of  the  air  particles 
whose  period  is  yjgth  of  a  second  and  which  produces  a 
certain  tone  usually  denoted  by  C.  If  the  same  note  is 
struck  on  the  piano,  it  is  found  that  the  principal  part  of  the 
motion  of  the  air  particles  again  has  y^-gth  of  a  second  as 
its  period.  But  the  motion  is  not  simply  harmonic.  It  is 
a  combination  of  this  fundamental  motion  with  one  twice  as 
fast,  with  another  three  times  as  fast,  and  so  on.  In  other 
words,  the  motion  of  the  air  particles  is  given  by  an  equation 
of  the  form 

(1)  y=.a^^m{^t^a^^-a^^m{-^t^a^ 

where  the  period  of  the  first  and  principal  term  is  T^  that  of 
the  second  ^  jT,  that  of  the  third  |  T^  etc. 

This  is  not  the  place  to  discuss  details  of  the  theory  of 
sound.  Our  purpose  in  entering  upon  this  theory  at  all 
was  merely  to  explain  one  of  the  many  instances  in  which 
sums  of  simple  harmonic  functions  of  the  form  (1)  present 
themselves  as  indispensable. 

We  wish  to  learn  how  the  various  terms  in  (1)  combine. 
For  that  purpose  the  length  of  the  period  jT  makes  but  little 
difference.     We  shall  therefore  put 

T=2  7r 
since   the  formulae   will  then  assume  a  somewhat   simpler 
appearance.     Then  (1)  reduces  to 

(2)  y  =  «!  sin  (t  +  ttj)  +  ^2  sin  (2  ^  +  Og) 

4- fl^s  sin  (3  ^  +  ctg)  +•••  . 
Now  each  of  these  terms  may  be  expanded  in  accordance 
with  the  addition  theorem  (Art.  79),  so  that 
ttj  sin  (t  -f  ttj)  =  a^  cos  a^  sin  t-\-  a^  sin  a^  cos  ^, 
^2  sin  (2  t  +  ttg)  =  ^2  ^os  H  si^  2  ^  4-  ^2  sin  a^  cos  2  t,  etc. 


GENERAL   HARMONIC   CURVES  241 

Consequently,  if  we  introduce  new  constants  J.^,  ^2'  "• '  -^v 
-^2'  •••  ^y  putting 

A^  =  a-^  sin  a^,    A^  =  ag  sin  Ogi  •  *  •  » 

j5j  =  a-^  cos  «!,    -Bg  =  ^2  ^^S  «2i  •  •  •  9 

equation  (2)  becomes 

.^                ^  =  ^2  cos  <4-^  cos  2  f  +  ^3  cos  3  i+ ••• 
+  B^  sin  ^  H-  ^2  si^  2  «  +  ^3  sin  3  «  H . 

Let  us  call  the  fixed  point,  with  which  the  moving  point 
would  tend  to  coincide  if  the  amplitudes  a^  a^<^  etc.,  of  all  the 
simple  harmonic  motions  of  (1)  were  to  approach  the  limit 
zero,  the  center  of  oscillation.  We  have  tacitly  assumed  so 
far  that  the  center  of  oscillation  was  the  origin  of  coordinates. 
Let  us  drop  this  specialization,  and  let  J  Aq  be  the  fixed 
value  to  which  y  would  reduce  if  all  of  the  oscillations  were 
to  disappear ;  that  is,  let  J  A^  be  the  ordinate  of  the  center 
of  oscillation.*  Then  we  must  add  ^  A^  to  the  right  member 
of  (3),  so  that  we  obtain  finally 

^A^          y  =  J  Aq  -h  ^1  cos  t-\-  A^  cos  2t  +  A^  cos  3  f  +  •  •  • 
+  ^1  sin  t-{-  B^^in2t^  B^  sin  3  i  H 

as  the  typical  equation  of  a  general  harmonic  motion. 

As  in  the  case  of  a  simple  harmonic  motion,  we  may  make  a  graphical 
record  of  this  motion.  Suppose,  for  instance,  that  the  motion  to  be  in- 
vestigated is  the  vibration  of  a  metallic  wire.  We  attach  a  light  pen  to 
the  wire  so  as  to  enable  it  to  write  upon  a  strip  of  smoked  glass.  If  the 
glass  be  left  at  rest  while  the  wire  is  caused  to  vibrate,  the  pen  will 
merely  describe  a  straight  line  upon  the  smoked  glass.  If,  however,  the 
glass  be  moved  with  a  rapid  uniform  motion  at  right  angles  to  the 
direction  of  vibration  of  the  wire,  there  will  appear  as  record  a  wavelike 
curve.     This  curve  will  belong  to  the  class  considered  in  the  next  article. 

109.  General  harmonic  curves.  If  we  put  x  in  place  of  t 
in  equation  (4)  of  Art.  108,  we  find 

(Vi        2^  =  2  ^0  +  ^1  ^^s  ^  +  ^2  ^^s  2x-\-  A^  cos  3  a;  4-  ••• 
H-  B^  sin  a;  +  -^2  sin  2  a;  +  ^3  sin  S  x  +  •  •  •  . 

*  The  reason  for  denoting  this  quantity  by  ^  Aq  rather  than  Aq  will  appear 
later.    (See  Art.  112.) 


242 


THEORY  OF   WAVE   MOTION 


The  curves  which  are  obtained  as  a  result  of  plotting  an 
equation  of  this  form  are  called  general  harmonic  curves  and 
are  capable  of  an  extraordinary  variety  of  forms.  In  fact  it 
can  be  shown,  by  methods  involving  the  integral  calculus, 
that  an  infinite  series  of  the  form  (1)  may  be  found  to  repre- 
sent almost  any  continuous  curve,  and  even  extensive  classes 
of  discontinuous  curves  *.  In  this  book,  however,  we  are  con- 
cerned only  with  sums  of  the  form  (1)  involving  a  finite 
number  of  terms  and  the  curves  represented  by  them.  The 
name  harmonic  curves  will  be  understood  to  apply  only  to 
such  curves. 

We  proceed  to  discuss  an  example.     Let  us  plot  the  curve   whose 
equation  is 

(2)  y  =  sin  a?  +  sin  2  ar. 
We  begin  by  drawing  the  two  familiar  curves 

(3)  y^  =  sin  x  and  7/2  =  sin  2  x, 

the  two  dotted  curves  of  Fig.  133.  From  these  curves  it  is  easy  to  con- 
struct the  curve  (2).     For  we  see  from  (2)  and  (3)  that 

y  =  2/1  +  2/2 
for  every  value  of  x.     If  then  we  find  the  ordinate  of  each  of  the  two 
dotted  curves  for  a  given  value  of  x,  their  algebraic  sum  will  be  the 
ordinate  of  a  point  on  the  required  curve.     The  resulting  curve  is  in- 
dicated in  Fig.  133  by  a  full  line.     A  few  points  of  this  curve  may  easily 


K 

/;''\  \"^N  /'  ^N         p. 

'i 

N 

+a; 


FiQ.  133 


*  Such  series  are  usually  called  Fourier's  series  in  honor  of  the  great  mathe- 
matical physicist  who  first  stated,  and  in  part  proved,  the  above  theorem.  The 
first  rigorous  proof  was  furnished  much  later  by  Dieichlbt. 


TRIGONOMETRIC   INTERPOLATION  243 

be  obtained  by  inspection.  For  x  =  0,  y^  and  7/^  are  both  zero,  and  there- 
fore also  y  =  y^-{-  y^  =  O.  Consequently  the  point  0  is  on  the  curve. 
For  X  =  7r/2,  yi  =  1  and  y^  =  0,  so  that  y=l  and  the  curve  passes  through 
the  point  A.  For  values  of  x  between  7r/2  and  tt,  y^  is  negative  so  that 
2/1  +  2/2  ^^i^l  b^  l^ss  ^^^'^  ^r  Consequently  the  full  line  curve  in  this  in- 
terval lies  below  the  corresponding  portion  ^jB  of  the  curve  y^  =  sin  x. 
For  a  certain  value  of  x  in  this  interval  (determined  by  the  equation 
sin  a:  -f  sin  2  a;  =:  0)  ^^  and  y^  will  be  numerically  equal  but  opposite  in 
sign,  so  that  at  that  point  the  curve  y  =  sin  x  -f  sin  2  a:  will  cross  the 
a;-axis.  This  is  the  point  L  of  Fig.  133.  For  any  value  of  x  we  may 
obtain  y^  and  y^  by  measurement  from  the  two  dotted  curves.  If  we 
form  the  sum  of  these  two  quantities  with  due  regard  to  sign,  we  find 
the  corresponding  ordinate  of  the  required  curve.* 


EXERCISE   LX 

Plot  the  following  harmonic  curves  : 

1.  y  =  2  sin  X  +  sin  2  x.  3.   y  =  sin  a:  -|-  cos  2  x. 

2.  y  =  sin  x  +  ^  sin  2  a:.  4:.   y  =  5  sin  x  +  sin  4  x. 

110.  Harmonic  analysis  or  trigonometric  interpolation.     We 

have  'seen  how  a  number  of  simple  harmonic  curves  may  be 
compounded  into  a  single  general  harmonic  curve.  It  often 
happens  that  a  curve  is  given,  actually  drawn  out  on  paper, 
as  for  instance  in  the  case  of  a  self-recording  barometer  or 
thermometer.  If  the  curve  is  of  a  periodic  character,  the 
question  arises  whether  it  may  be  regarded  as  a  harmonic 
curve,  that  is^  whether  it  is  possible  to  compound  it  out  of  a 
number  of  simple  harmonic  curves  by  the  method  of  Art. 
109.  And  if  so,  the  problem  presents  itself  to  actually  find 
the  component  simple  harmonic  curves.  The  process  of 
solving  this  problem  is  known  as  harmonic  analysis  and  is  of 
great  importance  in  many  branches  of  pure  and  applied 
mathematics. 

Let  us  suppose  that  the  given  curve  is  periodic,  so  that  it 


*  MiCHELSON  and  Stratton  have  devised  a  machine  for  performing  mechan- 
ically the  operation  of  combining  a  number  of  simple  harmonic  curves.  This 
machine  is  also  capable  of  performing  the  inverse  operation  discussed  in  Art,  110. 
For  this  reason  it  has  been  called  a  harmonic  analyzer. 


244 


THEORY  OF  WAVE   MOTION 


consists  of  an  infinite  number  of  equal  pieces,  and  let  the 
length  of  one  of  these  pieces,  the  wave  length,  be  equal  to  X. 
If  X  is  less  than  2  tt,  we  may,  by  a  process  of  magnification 
or  stretching,  replace  the  given  curve  by  another  one  similar 
to  it  whose  wave  length  is  just  equal  to  2  tt.  If  we  can  solve 
our  problem  for  this  second  curve  of  wave  length  2  tt,  it  will 
be  easy  to  solve  the  corresponding  problem  for  the  original 
curve.  If  X  is  greater  than  2  tt,  a  process  of  compression 
enables  us  to  reduce  the  problem  again  to  the  case  of  a  curve 
of  wave  length  2  tt.  We  may  therefore  assume  that  the 
wave  length  of  the  given  periodic  curve  is  equal  to  2  tt 
without  reducing,  by  this  assumption,  in  any  essential 
fashion  the  general  applicability  of  our  results.  The  ob- 
ject of  this  assumption  is  merely  to  simplify  the  resulting 
formulae. 

Let  PqP^P^  '"  (Fig.    134)   be  the  given   curve   of  wave 
length  OA  =  2  tt,  and  let  us  divide  OA  into  an  odd  number, 


+; 

J 

n^ 

-^^ 

0 

A 

V2 

2/3 

?A 

\ 

X,         X. 

A 

Vo 

/  ^' 

X2 

Xa 

X, 

\ 

f 

Po 

f 

\ 

% 

Ve 

A 

t 

i" ■ 

^ 

+« 


^a 


Fig.  134 


say  2  m  4- 1,  of  equal  parts.  Not  counting  A^  there  will  then 
be  2m-\-l  points  of  division;  namely,  0,  Xj,  X^,  •"  X^^^. 
In  Fig.  134  we  have  made  2m  +  l  =  l. 

At  these  points  of  division  we  construct  the  ordinates 
OPq.,  X-jP^^  JL2-P2'  *** 

of  the  curve.     Let  y^^  y^,  Vi-^"'  Vim  ^^  these  ordinates,  eacl 
with  its  proper  sign  prefixed.     On  account  of  the  periodic 
character  of  the  curve,  the  ordinate  at  A  will  be  the  same  as 
that  at  0.     This  is  the  reason  that  we  did  not  count  A  as 


TRIGONOMETRIC   INTERPOLATION  245 

one  of  the  points  of  division.  If  we  had  included  A^  we 
should  really  have  been  counting  0  twice. 

We  can  always  .find  a  harmonic  curve  involving  terms  in 
x^  2x,  Sx,  "'  mx,  Vvhich  passes  through  the  2  m  + 1  points 
Pq,  Pj,  Pg'  •••  ^2m'  ^^^^  *^®  general  equation  of  such  a 
harmonic  cuiive  is  » 

.^ .         ^  =  ^  Aq  +  A^  cos  x-\-  A^  cos  2x-\-  •  •  •  +  J.^  cos  mx 
-\-  B^sin  X  +  B2sin2x-\'  •  •  •  -|-  B^  sin  mx, 

and  therefore  contains  2m  +  l  coefficients  Aq,  ^j,  •••  ^, 
5j,  ••.  P^,  which  may  be  determined  in  such  a  way  as  to  make 
the  corresponding  curve  (1)  pass  through  the  2  m  +  1  given 
points.  In  fact,  the  curve  (1)  will  pass  through  the  point 
Pq,  if  the  value  of  y  obtained  from  (1),  for  a:  =  0,  is  equal  to 
the  ordinate  i/q  of  the  given  point  Pq  ;  that  is,  if 


(2)  I/O  =  1^0  4- A +  ^2+  -  +-^m. 

2  TT 

The  abscissa  of  P,  is  OX.  = .     Theref c 

^  ^      2m+l 

(1)  will  pass  through  Pj,  if  the  value  of  y  obtained  from 

(l),for:.=  -l^, 
2  m-h  1 

point  Pj ;  that  is,  if 


2  TT 

The  abscissa  of  P,  is  OX.  = .     Therefore  the  curve 

^  ^      2m+l 

irough  Pj,  if 

(1),  for  X  =  - — — — ,  is  equal  to  the  ordinate  «/,  of  the  given 
2  m-h  1 


27r       .     .    „„    2.27r 


(3)        ^1  =  1  ^0  +  ^1  cos — -  +  A  cos 


2m+l         ^        2w+l 

W  •  2  TT 


+  ...  +  J.^cos 


2m-\-l 


27r       .    „     .     2.27r 


H-^jSin- — ^^  +  P2sin 


2m+l         ^       2m+l 

W  .  2  TT 


+  ...  +^„sin 


2m  +  l 


In  the  same  way  we  find  that  the  curve  (1)  will  pass  through 
the  points  Pg,  Pg,  ••.  P^m  ^^  ^^®  following  additional  equations 
are  satisfied ; 


1 


246  THEORY  OF    WAVE  MOTION 

^    A     ,     A  47r       ,4  2  ■  47r    , 

2m-}-!        ^        2m  +  l        ^        2m  +  l 
,  ,    E»     .     m  •  4:7r 


(4) 


2w  +  l' 


2w4-l  _    2m4-  1  2w  +  l 

W  •  2  WTT 


+  ...  H-^^sin 


2w+l 


Now  the  equations  (2),  (3),  and  (4)  are  together  2  m  +  1  in 
number,  and  involve  2m +1  unknown  quantities,  viz.,  the 
quantities,  Aq^  A^  •  -  -  A^^  B^,  . . .  B^.  The  coefficients  of  these 
quantities  (sines  and  cosines  of  known  angles)  are  known 
numbers,  and  the  quantities  y^,  y^,  ...  ^^m')  ^^^  ordinates  of. 
the  given  points  Pq,  Pj,  ...  Pgm'  ^^®  known.  We  may  there-, 
fore,  in  general,  solve  these  2  m  +  1  equations  for  the  2  m  +  1; 
unknown  quantities  and  then  substitute  the  values  obtained, 
for  Aq,  ^j,  .••  A^,  B^,  ...  B^  in  (1).  The  harmonic  curve, 
obtained  by  plotting  (1)  will  then  pass  through  the  given 
points. 

If  the  given  curve  is  fairly  smooth,  and  if  the  points  Pq 
Pj,  ...  P^m  be  taken  close  enough  together,  that  is,  if  m  b< 
chosen  large  enough,  we  may  find  by  the  above  method 
harmonic  curve  which  may  be  regarded  as  replacing  th< 
original  curve  everywhere  with  any  desired  degree  of  aj 
proximation.* 

The  following  example  will  illustrate  this  method.  Let  y  he  a,  pei 
odic  function  of  x,  of  period  2  tt,  such  that  the  vajues  of  y  which  corre 
spoud  to 

X  =  0°,  72°,  144°,  216°,  288° 

*  A  more  precise  formulation  of  this  statement  must  be  left  for  a  much  latej 
stage  of  the  student's  mathematical  career. 


TRIGONOMETRIC   INTERPOLATION  247 

are  respectively 

y=Q,   +2,   +  i   -  i,  -  2. 

Our  general  theory  tells  us  that  we  may  find  an  expression  of  the  form 

y  =  \Aq-{-  ^icosa;  +  ^2 cos  2  a:  +  ^1  sin  a:  +  ^2  sin  2  x, 

which  assumes  the  five  values  assigned  to  y  for  the  five  given  values  of  x. 
Moreover  we  have  the  following  five  equations  for  the  five  unknown  co- 
efficients Aq,  Ai,  A2,  Bi,  B2'. 

Q  =  IAq  +  Ai  +  A2, 

2  =  ^  ^0  +  ^1  cos  72°  +  ^2  cos  144°  +  Bi  sin  72°  +  B2  sin  144°, 
I  =  ^  ^0  +  ^1  cos  144°  +  ^2  cos  288°  +  Bi  sin  144°  +  B2  sin  288°, 

-  1  =  ^  ^0  +  ^1  cos  216°  +  .42  cos  72°  +  Bx  sin  216°  +  B^  sin  72°, 

-  2  =  I  ^0  +  ^1  cos  288°  +  ^  2  cos  216°  +  ^1  sin  288°  +  B2  sin  216°. 

Since  we  have 

sin  216°  =  sin  (360°  -  144°)  =  -  sin  144°, 
cos  216°  =  cos  (360°  -  144°)  ^  cos  144°, 
sin  288°  =  sin  (360°  -  72°)  =  -  sin  72°, 
cos  288°  =  cos  (360°  -  72°)  =  cos  72°, 

the  above  equations  may  also  be  written  as  follows  : 

(1)  J  ^0  +  ^1 +  ^2-0, 

(2)  \  Ao  +  ^1  cos  72°  +  ^2  cos  144  +  5^  sin 72°  +  ^3  sin  144°    =  2, 

(3)  ^  ^ 0  +  ^  1  cos  144°  +  ^ 2  cos  72°  +  B^  sin  144°  -  B^  sin  72°    =  |, 

(4)  \Aq  +  A^  cos  144°  +  ^2  cos  72°  -  Bi  sin  144°  +  B2  sin  72°  =  -  1, 

(5)  \Aq  +  Ai  cos  72°  +  Ai  cos  144°  -  Bx  sin  72°  -  Bi  sin  144°  =  -2. 

From  (2)  and  (5)  we  find  by  addition 

(6)  ,\Aq  +  Ax  cos  72°  +  A2  cos  144°  =  0, 
and  similarly  from  (3)  and  (4), 

(7)  ^  ^0  +  ^1  cos  144°  +  J  2  cos  72°  =  0. 

From  (1)  we  have 

^4o  =-  ^1  -  ^2 

which,  substituted  in  (6)  and  (5),  gives 

(8)  Ax  (cos  72°  -  1)  +  ^2  (cos  144°  -  1)  =  0, 

Ax  (cos  144°  -  1)  +  ^2  (cos  72°  -  1)  =  0. 

If  we  multiply  both  members  of  the  first  of  these  equations  by  cos  72°  —1, 
those  of  the  second  by  —  (cos  144°  —  1),  and  add,  we  find 

(9)  ^i[(cos  72°  -  1)2  -  (cos  144°  -  1)2]  =  0. 

From  the  table  of  natural  functions,  we  find  to  two  decimal  places 
cos  72°  =  0.31,  cos  144°  =  -  cos  36°  =  -  0.81. 


248  THEORY  OF  WAVE  MOTION 

Therefore 

cos  72^  -  1  =  -  0.69,  cos  144°  -  1  =  -  1.81, 

so  that  the  coefficient  of  u4j  in  (9)  is  not  equal  to  zero.  Consequently 
we  conclude  from  (9)  that  A^  =  0.  According  to  (8)  and  (1),  we  must 
then  have  also  ^.j  =  0,  ^^  =  0. 

If  now  we  put  Af^=  A^  =  A^^  0,  in  (1)  to  (5),  these  five  equations 
reduce  to  the  following  two  : 

B.  sin  72°  +  B„  sin  144°  =  2, 

rio)  1  -T-    2 

^     ^  B^  sin  144°  -  B^  sin  72°  =  |. 

From  these  equations  we  eliminate  first  B^  and  then  B-^^ ,  giving 

(sin2  72°  +  sin2144°)5i  =  2  sin  72°  +  |  sin  144°, 

^     ^  (sin2  72°  +  sin2 144°) ^3  =  2  sin  144°  -  \  sin  72°. 

From  the  table  of  natural  sines  we  find,  correct  to  two  decimal  places, 

sin  72°  =  0.95,  sin  144°  =  sin  36°  =  0.59, 
so  that 

sin2  72°  +  sin2  144°  =  0.90  +  0.35  =  1.25. 

Consequently,  equations  (11)  become 

1.25^1=  1.90  +  0.30  =  2.20, 

1.25i?2  =  1-18 -0.48=:  0.70; 
whence  finally 

(12)  ^1=1.76,  ^2  =  0.56. 

Since  we  have  already  found  Afy  =  A^=  A^-=  0,  the  function  which  we 
were  seeking  is 

(13)  y  =  1.76  sin  a;  +  0.56  sin  2  a;. 

In  order  to  check  our  result  we  may  substitute  the  five  given  values  of 
X  in  (13)  and  verify  that  the  corresponding  values  of  y  are  actually  those 
which  were  originally  given.  That  equation  (13)  gives  y  =  0  for  a:  =  0  is 
obvious.     For  x  =  72°  and  for  x  =  144°,  we  find  from  (13) 

y  =  1.76  X  0.95  +  0.56  x  0.59  =  1.67  +  0.33  =  2.00, 
and 

y  =  1.76  X  0.59  +  0.56  x  (-  0.95)  =  1.04  -  0.53  =  0.51, 

respectively,  checking  to  within  one  unit  of  the  last  decimal  place 
computed.  To  check  the  other  two  pairs  of  values  requires  no  additional 
computation. 

In  this  example,  the  values  of  y  iov  x  =  144°  =  180°  —  36° 
and  for  aj=  216°  =  180°  +  36°  were  numerically  equal  but 
opposite  in  sign.  The  values  of  y  iov  x=  72°  =  180°  —  108° 
and  for  x  =  288°  =  180°  +  108°  were  also  numerically  equal 


TRIGONOMETRIC   INTERPOLATION  249 

but  opposite  in  sign.  It  is  owing  to  this  circumstance  that 
J.Q,  J-i,  and  A^  turned  out  to  be  all  three  equal  to  zero. 
Having  become  aware  of  this  fact,  we  may  abbreviate  our 
work  very  much  by  at  once  equating  Aq,  A^,  A^,  etc.,  to  zero, 
whenever  the  given  values  of  the  function  are  numerically 
equal  but  opposite  in  sign  for  those  among  the  given  angles 
which  differ  numerically  by  the  same  amount  from  180°. 

Similarly,  if  the  given  values  of  the  function  y  are  equal 
numerically  and  in  sign  for  all  of  those  among  the  given 
angles  x  which  differ  numerically  by  the  same  amount  from 
180°,  the  expression  for  7/  will  contain  no  sine  terms  ;  that  is, 
^j,  B^t  -Sg,  etc.,  will  all  be  equal  to  zero. 

EXERCISE     LXI 

1.  Find  a  periodic  function,  involving  only  the  angles  x  and  2  x, 
which  assumes  the  values 

2/  =  0,  +  2,  +  1,  -  1,  -  2, 
for  X  =  0°,  72°,  144°,  216°,  288°, 

respectively.     Compute  the  coefficients  to  two  decimal  places. 

2.  Find  a  periodic  function,  involving  only  the  angles  x  and  2  x, 
which  assumes  the  values 

2/  =  +  2,  +  1,  -  i,  -  i  +  1, 
for  x  =  0°,  72°,  144°,  216°,  288°, 

respectively.     Compute  the  coefficients  to  two  decimal  places. 

111.  Theorems  leading  to  the  general  solution  of  the  problem 
of  trigonometric  interpolation.  We  have  shown  in  Art.  110 
that  the  problem  of  trigonometric  interpolation  may  be  re- 
duced to  that  of  solving  a  system  of  2  m  +  1  equations  of  the 
first  degree  with  2m-\-  1  unknowns.  But  we  can  accomplish 
much  more  than  this.  We  shall  derive  elegant  and  con- 
venient formulae  for  the  solutions  of  these  equations,  en- 
abling us  to  find  the  values  of  the  coefficients  Ak  and  B^  by 
a  direct  and  simple  process.  But  before  we  can  do  this,  we 
must  prepare  the  way  by  proving  some  theorems  necessary 
for  this  purpose. 


250  THEORY  OF  WAVE  MOTION 

We  begin  by  proving  that  the  following  formula 
(1)    sina  +  sin(a  +  0  +  sin(«  +  2  04-  •••  -{-s'm(a  +  mt) 

/     .  mt\   .    (m-\-  l)t 
sm(  a  +  —]sm^ — ^|— ^ 


.     t 
sm- 

published  by  Euler  *  in  1743,  is  true  for  all  values  of  a  and  t, 
provided  that  sin  -  is  not  equal  to  zero. 

Proof.  Let  us  denote  the  sum  in  the  left-hand  member 
of  (1)  by  8^.     If  we  multiply  ^^  by  2  sin  -,  we  shall  have 

A 

2  s„  sin  -  =  2  sin  a  sin  -  +  2  sin  (a  +  t)  sin  - 

A  ''A  A 

+  2  sin  (a  +  2  t)  sin  -  +  •••  +2  sin  (a  +  mt)  sin  |. 
A  A 

Every  term  in  the  right  member  of  this  equation  contains 
a  product  of  two  sines,  and  may  therefore  be  expressed  as  a 
difference  of  two  cosines  by  means  of  formula  (4)  of  Art.  82  ; 
that  is, 

sin  a  sin  yS  =  I  [cos  (a  —  yS)  —  cos  (a  +  /S)]. 

We  find,  in  this  way, 
2  8^  sin  -  =  cos  {a  —  1 0  ~  ^^^  (^  +  i  0 

A 

+  COS  {a  4-  \  0  —  <JOs  (a  +  I  0 

+  cos  (a  +  I  ^)  —  cos(a  H- f  0 
+ 

+  cos  5  61  +  (w  —  \)t\  —  COS  \a-\-{m-\-  \)t\ . ' 

*  EuLER  (1707-1783)  was  born  in  Switzerland,  but  spent  most  of  the  years  of 
his  scientific  career  in  St.  Petersburg  and  Berlin.  His  work  was  of  fundamental 
importance  in  all  parts  of  pure  and  applied  mathematics.  Although  absolutely 
blind  during  the  latter  part  of  his  life,  he  continued  to  labor  and  to  make  im- 
portant contributions  up  to  the  end. 


TRIGONOMETRIC  INTERPOLATION  251 

Clearly  all  of  the  terms  in  the  right  member  except  the 
first  and  last  will  destroy  each  other,  so  that  we  are  left 
with  the  equation 

2  «^sin  -  =  cos  (a  —  J0"~  cosja  -\-(m-\-  |)^j. 

A 

But  we  have  the  formula  (see  Art.  82,  equations  (5)), 

.  DO-    A-B       A-\-B 

cos  A  —  cos  B  =  —  2  sin  — - — cos  — ^ — , 

SO  that 

2  «„  sin  -  =  —  2  sin  -  (  —  ^  —  ^0  ^^^  o  (^  ^  +  ^0 

A  A  A 

=  2sinU  +  — jsin^^ — 2^' 

whence,  if  sin  -  is  not  equal  to  zero, 

z 

This  is  the   same  as  equation  (1),   the  formula  which  we 
wished  to  prove. 

Let  us  rewrite  formula  (1)  with  a'  in  place  of  a,  and  then 
put 

Since  sin  a' =  sinf  a +  ^]  =  cos^, 

we  then  find 

(2)     c^=cos  a-\-  cos  (a  +  t)-\-  cos(a+  2  t)-\ —  +  cos(a  +  mt) 


(     ,  rnt\    .     0 


cos(a  +  ^lsin->±l^ 


.    t 
sin- 


252  THEORY   OF   WAVE   MOTION 

a  formula  which  may  also  be  obtained  directly  by  a  process 
strictly   analogous  to  the  one  employed   for   the  proof   of 
equation  (1).     Formula  (2)  is  also  due  to  Euler. 
If  in  equations  (1)  and  (2)  we  put  a  =  0,  we  find 

sin  —  sin  ^^ — - — — 
(3)     sin  f+sin  2  ^+ sin  3  ^H +^\nmt  = 


and 


sm^ 


cos  --  sin  -^^ — '     ^ 

2  2 


(4)  l-|-cosf  +  cos2f/H-cos3fH \-co^mt  = -^ * 

sin- 

Let  us  subtract  J  from  both  members  of  (4).     We  obtain 
the  formula 

mt   .    (m  4-1)^ 
cos  —  sin  -^^ ' — — 

1  2                2             1 
-  +  cos  1 4-  cos  2  ^  H h  cos  mt  = 

^  '      t  A 

sin- 

o         mt   .     (m-^l)t        .     t        .    f      ,  1\^  ,     .    t        .    t    ' 

2  cos  —  sin       ^    ^ sm  -      sin [m-i--Jti-  sin- -  sin- 

2  sin  -  2  sin  - 

2  2 

where  we  have  made  use  of  one  of  the  equations '(4)  of  Art< 
82.  The  numerator  of  the  last  fraction  obviously  reduces  ti 
its  first  term,  so  that  we  find  finally 

sin  ^^ -! — ^ 

1  2 

(5)  -  +cosf+cos  2*+cos  St-i —  -\-cos  mt  = r 

2  2  sin  - 
a  formula  which  was  known  to  Snellius  in  1627. 


The  angle  t  which  occurs  in  all  of  these  equations  ma^ 
have  any  value  whatever  excepting  only  those  values  foi 

which  sin  -  is  equal  to  zero ;   that  is,  excepting  the  valueJ 

Li 

2  Ajtt,  where  h  is  an  integer.     We  shall  now  apply  these  f  ormuh 


TRIGONOMETRIC   INTERPOLATION  253 

to  the  particular  case  when  ^  is  a  commensurable  fractional 
part  of  the  entire  circumference,  so  that 

n 

where  both  h  and  n  are  positive  integers  and  where 

n>l. 

We  shall  further  put        m  =  n—\. 

We  may  express  these  assumptions  more  concretely  as  fol- 
lows.    Let  us  divide  the  circumference  of  the  circle  into  n 

equal  parts,  where  n>  1.  Then  -^  will  be  the  angle  sub- 
tended by  one  of  these  parts.  The  smallest  multiple  of  this 
angle  which  is  equal  to  a  complete  circumference  is  of  course 
the  nth..     Therefore  the  angles 

iz,  2?^,  3-^,...(ri-l)^ 
n  n  n  n 

are  all  distinct,  and  we  propose  to  find  the  sum  of  their  sines 
as  well  as  of  their  cosines.     We  consider  next  the  angle 

2 = and  all  of  its  multiples  up  to  (w  —  1) and  calcu- 

n         n  n 

late  the  sum  of  their  sines  and  of  their  cosines.     In  general, 

we  consider  the  angle  k  ^^^  =  - — -  and  its  multiples  2 — —, 

n  n  n 

S- — —'"  (ri  —  1)  — —  and  compute  the  sum  of  their  sines  and 
n  n 

of  their  cosines. 

According  to  (3)  we  have  (putting  t  = and  m  =  n  —  l') 

sm h  sm  2 \-  sm  3 h  •  •  •  -I-  sm  (w  —  1) 

n  n  n  n 

.     (n—  l)k7r   .     J 

sm  -^^ '^ —  sin  fCTT 

n 


.     kir 

sm  — 

n 

a  formula  which  will  be  valid  unless  sin  — is  equal  to  zero; 


254  THEORY  OF   WAVE  MOTION 

that  is,  unless  k  is  an  integral  multiple  of  n.  Excluding 
this  case  for  a  moment,  we  see  that  the  right  member  is  equal 
to  zero  since  sin  kir  occurs  as  a  factor  in  the  numerator  and 
since  the  sine  of  any  integral  multiple  of  ir  is  equal  to  zero. 
Consider  now  the  excluded  case  when  ^  is  a  multiple  of  n. 
Then  every  term  on  the  left  member  is  individually  equal  to 
zero.     We  see  therefore  that 

(b)     iSj^  =  sm f-  sin  2 f-  sin  3 h  •  •  • 

n  n  n 

+  sin  {n  —  1) =  0, 

n 

for  every  value  of  h^  whether  k  is  divisible  by  n  or  not. 

We  may  prove  in  the  same  way,  making  use  of  equation 
(4),  that 

^r7^  />       -I    ,         2  kTT  ,  o  2  kir  ,  o  2  ^TT  , 

(7)  Ch=l-hcos |-cos2 |-cos3 (-  ••• 

n  n  n 

+  cos  (n  —  1) =  0, 

n 

ifk  is  not  a  multiple  of  n. 

If  A;  is  a  multiple  of  n,  all  of  the  angles ,  2- ,  etc., 

n  n 

are  integral  multiples  of  2  tt,  so  that  each  of  the  cosines  which 

appears  in   OJ.  is  equal  to  unity.     There  are  n  terms  in  Cj,. 

Therefore 

(8)  Ok=  1  +  cos h  cos  2 h  ...  4-cos  (n—l) =n, 

n  n  n 

if  k  is  a  multiple  of  n. 

2  kir  2  /tt 

Let  us  now  consider  two  angles  of  the  form and , 

n  n 

and  denote  by  Oki  the  sum 

(9)  Cki—i^-\-  cos cos h  cos 2 cos  2 

n  n  n  n 

+  •••  +  cos  (n—  1) cos  (n  —  1) 

^  n  n 


TRIGONOMETRIC   INTERPOLATION 


255 


A  product  of  two  cosiaes  may  be  expressed  as  a  sum  by 
means  of  formula  (3)  of  Art.  82  ;  that  is, 

cos  a  cos  y8  =  |^[cos  (a  —  /3)  +  cos  (a  +  y8)] . 

Therefore  we  may  write,  if  we  apply  this  formula  to  each 
term  of  C^i^ 

[_  n  n        J 


+  J  COS 2—^ ^—  +  cos2   ^    ^   ^ 

[_  n  n        ^ 


+ 


+  I  rcos  (n  -  1)  ^(^  -  ^)"  +  cos  (n  -  1)  ^(^  +  0^ 

Now  the  sum  of  all  of  the  terms  in  the  first  column  may  be 
equated  to  \Ck-i\i  we  again  make  use  of  the  notation  0^  de- 
fined by  equations  (7)  and  (8).  Similarly  we  observe  that  the 
sum  of  the  terms  in  the  second  column  is  ^  C^^i.     Therefore 

(10)  (7«  =  KC.-,+  <7w)- 
Consider  now  the  expression,  analogous  to  (7^^, 

(11)  ^Ski  =  sin sin f-  sin  2 sin  2 1-  •  •  . 

n  n  n  n 

+  sin  {n  —  1) sin  (n  —  1) • 

n  n 

Since  we  have  (see  Art.  82) 

sin  a  sin  y8  =  1^  [cos  (a  —  /3)  —  cos  (a  +  y8)], 
we  find 


^ki 


H[. 


<?.,=  i    COS 


2(k-l)7r_^2(k 


COS 


n 


n        J 


,    if        ^^Ck-l^TT  o2(^  +  Z)7r"| 

+  \\  COS  2—^^ ^- COS  2—^^ — — ^— 

\  n  n        ] 

+ 

+  iLos  (.  -  l)?i^::iQ^  -  cos  (.  -  l)K^±il5, 


256 


THEORY  OF  WAVE  MOTION 


so  that 

(12) 

Sm  = 

KO.-i- 

0,^d' 

Finally 

let 

us 

put 

(13)   (S,, 

Oi) 

= 

sin  2^- 
n 

2l7r 

cos 

n 

+  sin2 

2A;7r 
n 

cos  2 

2lir 
n 

+ 

sin  (n  - 

n 

cos  (n  - 

-1) 

2lir 
n 

+ 


Since  we  have  (see  Art.  82) 

sin  «  cos  yS  =  \  [sin  («  —  /S)  +  sin  (a  +  /3)], 
we  find  by  a  repetition  of  the  above  method 

(14)  (^.,CO=J('S'*-,  +  '^w)- 

We  have  already  shown  (cf.  equations  (6),  (7),  and  (8)) 
that 


(16) 


S^=^  for  all  values  of  ^, 

Ok  =  0  for  all  values  of  k  which  are  not  divisible  by  n, 

Ck  =  n  for  all  values  of  k  which  are  divisible  by  n. 


If  we  make  use  of  these  facts,  equations  (10),  (12),  and 
(14)  now  teach  us  that  the  following  statements  are  true. 


(16) 


(17) 


Oki  =  0  if  neither  k  —  I  nor  k+  I  is  divisible  by  n^ 

^ki  =  o  ^^  either  k  —  I  or  k  -\- 1  is  divisible  by  n,  but 

not  both, 
Ojci  =  n  ii  both  k  ^  I  and  k  -\-l  are  divisible  by  n. 

Ski  =  0  if  neither  k  —  I  nor  ^  +  Z  is  divisible  by  n, 

^ki  =  +  Q  i^  k—l  is  divisible  by  n  and  k  +  1  is  not 
divisible  by  /i, 

>^«  =  —  ^  if  k—l  is  not  divisible  by  n  and  k  +  1  is 

divisible  by  w, 
Sf^i  =  0  il  k  —  I  and  k  -}- 1  are  both  divisible  by  n. 
(^Sk,  Oi)  =  0  for  all  values  of  k  and  l. 


TRIGONOMETRIC   INTERPOLATION  257 

112.  General  solution  of  the  problem  of  trigonometric  inter- 
polation. We  are  now  ready  for  the  solution  of  the  problem 
of  trigonometric  interpolation.  We  divide  the  circumference 
into  2  m  H-  1  equal  parts  by  means  of  the  angles 

—  0  2  TT  47r  6  TT  4  mir 

'    2m  +  l'    2m+1'    2m  +  l'"''    2m  +  l' 
and  denote  the  corresponding  values  of  the  function  y  by 

^0'     VV     Vv     ^3'    •••  V'hm 

respectively.     The  function 

y  =  \A^-\-  A-^  cos  x-{-  A^  cos  2x-\-  •  •  •  +  ^,„  cos  mx 
+  j5j  sin  2;  +  ^2  si^^  ^  ^  +  •*•  +  -^m  sin  wa; 
will  actually  assume  these  values,  if  the  2  w  +  1  quantities 
Aq,  J.J,  ...,  J.^,  5j,  ^2^  "'1  ^m  ^1*6  chosen  so  as  to  satisfy  the 
2  m  4- 1  equations 

3/0  =  2  A  +  -^1  +  A  +  •••  +  -^m. 

0  2  TT 

^i=2A  +  Acos- — -+   ...  +J.^cosm 


(1) 


2m  +  l  "*  2m  +  l 

+  Asin- — -+  •••  +^mSinm- — -, 

2mH-l  2mH-l 


^2m  =  2  A  +  A  cos — -  +  ...  +  An  cos  m 


2m+l  2m+l 

+  A  sin  7^ —- +  •••  4-^^sin  w- — , 

2m-\-l  2  m-^1 

all  of  which  may  be  expressed  by  the  single  equation 
(2)        yv  =  i  A  +  A  cos —-  +  •••  +  A  cos  m 


2m+l  2m+l 

2  I/TT  .  .      T^         .  2  Z^TT 


H-  ^1  sin 4-  •••  +  B^  sm  m 


2m  +  l  2w-hl 

if  we  think  of  v  as  assuming  in  succession  the  2  w  +  1  values 
1^  =  0,  1,  2,  3,  ...,  2m. 
Let  r  be  any  integer   between  0  and  m,  let   us  multiply 

Vq   oy  1,     ^1  by  cos  — -,    y^   by  cos  2 -— ,   ...,    y^  by 

m  +  1  772  +  1 


258  THEORY  OF   WAVE  MOTION 

cos  V -,  ...,  «/2^  by  cos  2  m  —^ -,  and   add.     If  we 

m  +  1  2m  +  1 

make  use  of  equations  (1)  and  the  notations  introduced  in 

Art.  Ill,  we  find 

(3)  y^+y^00S^^+  ...  +2,,„C0S2™^|^ 

From  Art.  Ill,  (17),  we  know  that  all  of  the  quantities 
(^Sk^  Ci)  are  equal  to  zero.  Since  r  was  chosen  as  an  integer 
between  0  and  w,  r  can  be  divisible  by  2?7i  +  1  (which  num- 
ber corresponds  to  the  n  of  equations  (15),  (16),  (17)  of 
Art.  Ill),  only  if  r  =  0.  In  that  case  we  shall  have,  accord- 
ing to  (15)  of  Art.  Ill, 

Or=  0^  =  71=- 2m +  1, 

and  according  to  (16),  Art.  Ill, 

^10  ~  ^20  =   •••   =  Wio  ~  ^» 

since  none  of  the  numbers  1,  2,  •••,  m  are  divisible  by  2m-\-l. 
Consequently,  equation  (3)  reduces  to 

(4)  ^0  +  yi  +  ^2  +    -  +  y2m  =  Mo(2  ^  +  1) 

in  the  case  r=0. 

If  r>0,  Cr=0.  Moreover,  (7^  will  be  zero  for  all  values 
of  k  for  which  neither  k  —  r  nor  k  -\-r  is  divisible  by  2  w  -h  1. 
But  ^  as  well  as  r  can  at  most  be  equal  to  m,  so  that  neither 
k  —  r  nor  ^  +  r  is  ever  as  large  as  2m-\-l.  The  only  case 
therefore  in  which  one  of  these  numbers  can  be  divisible  by 
2  m -hi  is  when  k  —  r.  Thus  we  have,  according  to  (16), 
Art.  Ill, 

Ckr  =  0  iov  k  different  from  r, 

^  _2m-hl^ 


TRIGONOMETRIC   INTERPOLATION  259 

Consequently  equation  (3)  gives,  for  r  >  0, 

(5)    ^'o  +  j/icos^— ^  +  y,cos2  2^^^+... 

o         2r7r         2m  +  l   . 

+  Vo^  cos  2  m = — ! —  A.^ 

^2'"  2w  +  l  2 

and  we  notice   that   equation   (4)   may  be   thought    of  as 
included  in  (5)  for  r  =  0.*     We  find  therefore  the  formula 

2  7-7r      . 


('^  ^^=2^^+^^^^^rirTi+^^'"'' 


2w  +  l 

+  Vorr.  cos  2  m - 

(r  =  0,  1,  2,  ...,7/i), 

enabling  us  to  compute  A^,  A^^  ••-,  J.^  in  terms  of  the  given 
quantities  ?/o,«/i,  -",  y^^. 

It   remains  to  find  a  corresponding  formula  for  B^.     In 
order  to  do  this  we  return  to  equations  (1),  multiply  them 

m  order  by  0,  sm- -^  sin  2- -,...,  sinzw -, 

and  add.     This  gives 

^ism- -— +  ?/2Sin2- —}■  +  •••  +  y2mSm2w — 

2  m  -\-  1  2  m  -\-  1  2m  +  l 

=  i^.S',  +  A,(S,,  Ci)  +  A(^..  ^2)  +  -  +  An(^.,  6U 

+  A^lr  +  ^2^2r  H ^^  ^m^mr' 

All  of  the  terms  in  the  right  member  of  this  equation  reduce 
to  zero  except  B^S^^^  which,  according  to  Art.  Ill,  (17),  be- 
comes equal  to  B^    ^    —    Consequently  we  find 


a)    ^.= 


2m  +  l 


Vi  sm- +  v„  sin  2 -f-  •  •  • 

^^       2m +  1^^  2m +1 

•     o         2r7r   1 
+  ^2.sm2m^--^J 

(r=l,  2,  3,  ...,w). 


*  It  was  for  this  reason  that  the  notation  5  ^0  >  rather  than  Aq  ,  was  chosen  in 
Art.  108  for  the  constant  term  of  the  harmonic  fmiction. 


260  THEORY   OF   WAVE  MOTION 

Equations  (6)  and  (7)  furnish  the  complete  solution  of 
the  problem  of  trigonometric  interpolation.  The  coefficients 
of  the  harmonic  function 

1/  =  IAq-\- A^Gosx-{- A^cos2x-\ —  -\-A^Gosmx 
+  B^  sin  X  -{-  B^  sin  2  X  -^  •  '•  -{-  B^  sin  mx 

which  assumes  the  arbitrarily/  assigned  values 

for  the  2  m  -{-1  equidistant  values  of  the  argument 

—  0          ^  TT              4  TT  imir 

^~    '    2m+l'     2w  +  l' '  2m+  1' 

are  given  hy  equations  (6)  and  (7). 

EXERCISE  LXII 

1.  Solve  the  problem  treated  in  detail  in  Art.  110  by  the  method  of 
Art.  112. 

2.  Solve  the  problems  of  Exercise  LXI  by  the  general  method  of 
Art.  112. 


INDEX 

References  are  to  pages 


Abscissa 

Accuracy,    degree   of,    in    cal- 
culation       

of  five-place  tables .... 

of  measurements    .... 
Addition  theorems       .   184,  185, 
Ambiguous  case  of  oblique  tri- 
angles      Ill- 
Amplitude      228, 

Analysis,  Harmonic   .... 
Angle,  betw^een  directed  lines 

cardinal 145, 

general 

horizontal 

inclined 

initial  side  of 

measurement  of      .... 

natural  measure  of      .     .     . 

negative 

of  elevation  or  depression    . 

positive 

quadrant  of  an 

standard  position  of    .     .     . 

subtended 

terminal  side  of      .... 

vertical 

Arc,  length  of 

functions  of 

5  Area  of  a  triangle  .     .     .    84,  91 

Aristotle 

^  Aryabhata 155, 

Auxiliaries  S  and  T     .     .        61, 
Axis  of  a  coordinate  system 

rBall       .    •. 

Base  of  a  system  of  logarithms 
Bearing 


138 

37 

59 

1-5 

212 

114 
232 
243 
204 
173 
133 

73 

73 

135 

3 

165 

134 

73 
134 
145 
136 

74 
135 

73 
165 
152 
,92 

47 
192 
198 
138 

47 
43 
74 


Briggs 47 

Biirgi 38 

Cajori 39 

Center  of  oscillation    ....    241 

Chaining 74 

Characteristic,      of      a     loga- 
rithm      47,  49 

Checks 30,  70 

Chord 155 

Cof  unctions 21 

Cologarithm 54 

Compass 5,  74 

Complementary    angles,   func- 
tions of 20,  169 

Composition,  of  displacements, 
velocities,  forces  and  vectors     126 

Coordinates 138,  139 

Cosecant,  definition  of     .     .   14,  140 

graph  of 163 

line  representation  of .     .     .     154 

Cosecant  Curve 163 

Cosine,  definition  of   .     .  14,  87,  140 

graph  of 161 

line  representation  of .     .     .     152 

Cosine  Ciurve 161 

Cosines,  law  of 85 

Cos  (aiP) 184,186 

Cos  A  ±  cos  5 188 

Cos  2  a 189 

Cotangent,  definition  of  .     .    14,  140 

graph  of 163 

line  representation  of      .     .     153 

Cotangent  Curve 163 

Cot  (a  ±  p) 185,  186 

Crelle's  Tables 38 

Crest  of  a  wave 232 


261 


262 


INDEX 


Decimal  places,  number  of  .     .  34 

Departure 75 

Descartes 130,  159 

Difference  of  latitude       ...  75 
Difference  of  two  angles,  func- 
tions of 186 

Difference  of  two  cosines      .     .  188 

of  two  sines 188 

Directed  lines 201 

line  segments 202 

Direction  cosines 209 

Dirichlet 242 

Displacements 126 

Distances,  measurements  of     .  1 

Double  angles,  functions  of      .  189 

Eichhorn  slide  rule      ...     66,  95 
Equations,  trigonometric       222-225 

Errors,  gross 68 

small 70,  71 

Eudemus 7 

Euler 250 

Extraction    of    roots  by  loga- 
rithms     55 

Fermat 159 

Fink 101 

Forces 127 

Form  ratios  of  a  triangle     .     .      98 

Foster 63 

Fourier 242 

Framework  of  a  calculation      .      69 

Fuller's  slide  rule 65 

Function,  definition  of    .     .     .       13 

graph  of 156-163 

Functions,  of  acute  angles  .  14 
of  cardinal  angles  .  .  .  147 
of  complementary  angles  20,  169 
of  double  angles  .  .  .  .  189 
of  a  general  angle       .      137,  140 

of  half  angles 190 

of  obtuse  angle       ....       89 
of    small   angles   or  angles 
near  a  right  angle   ...       60 


of  sum  or  difference  .      184,  186 

of  supplementary  angles     .  89 

of  symmetric  angles  .     .     .  167 

Grade 12 

Gradient 12 

Graphic  method       .....  8 
Graphs,  of  functions,  some  of 

whose  values  are  given     .  156 
of    simple   algebraic    func- 
tions         158 

of  the  trigonometric   func- 
tions              159-163 

Greek  alphabet 132 

Gross  errors 68 

Gunter  . 62 

Gunter's  scale     ......  61 

Half  angle  formulae    ....  95 

Half  angle,  functions  of      .     .  190 

Harmonic  analysis 243 

Harmonic  curves,  simple      .     .  231 

general 241 

Harmonic  motion,  simple     .     .  226 

general 239 

Heights  and  distances      .     .     .  118 

Hero 92 

Hero's  formula 92 

Hipparchus 192 

Horizontal    angles,    lines  and 

planes 73 

Identities,  involving  a  single 

angle 149 

involving  several  angles       .  185 

Inaccuracies  of  measurements  .  1 
Inclined  lines,   planes   and 

angles 73 

Index  of  Refraction    .     .     .     .  129 

Index  laws 41 

Indices,    fractional,   negative 

and  zero 40 

Inskip's  Tables 31 


INDEX 


263 


Interpolation,  in  use  of  tables.      26 

trigonometric 243 

Inverse  functions 217 

Latitude 78 

Law,  of  cosines 85 

of  sines      ....       84,  96,  212 

of  sines  generalized    .     .     .  212 

of  tangents 101 

Level 5 

Level  chaining 80 

Leveling 74 

Limits  of  variation  of  the  func- 
tions         147 

Limits  i^^  and  *^      ...     195 

e  e 

Line  representation    of    func- 
tions        151 

Lines,  directed       201 

Line  segments,  directed  .     .     .    202 
Logarithms,  arrangement  and 

use  of  the  tables  .  .  51-53 
base  of  a  system  ....  43 
calculation  with  ....  47 
characteristics  of    .     .     .     47,  49 

common 47 

definition  of 43 

extraction  of  roots  by     .     .       55 

mantissa  of 47,  48 

of  numbers 47-56 

of  trigonometric  functions        57 
properties  of      ....      44, 45 
Logarithmic     calculations    in- 
volving negative  numbers       56 

Logarithmic  scale 61 

Longitude 78 

Mannheim 64 

Mannheim's  slide  rule      ...  63 
Mantissa       .(....     47,48 

Mariner's  compass       ....  75 
Measurement    of    lines    and 

angles 1,  3 

Menelaus        192 


Meridian 74 

Method  of  mathematical  induc- 
tion     170,  184 

Michelson 243 

Mobius 213 

MoUweide 105 

MoUweide's  equations     .     .     .  104 

Napier       38 

Nasir  Addin 84 

Natural  functions 25 

Natural  unit  of  circular  meas- 
urement         165 

Navigation 75 

Negative  angles 134 

line  segments 202 

Newton 105 

Object  of  trigonometry  ...  10 
Oblique   triangles,   ambiguous 

case  of 111-114 

area  of 84,  91,  92 

solution  of 107-131 

theory  of 82-106 

Ordinate 138 

Origin   of   arcs,   primary  and 

secondary 151,  152 

Origin  of  coordinates  ....  138 
Oscillation,  center  of  ...  .  241 
Oughtred 62 

Parallax 198 

Parallelogram  of  forces  .  .  .  127 
Periodicity     of     trigonometric 

functions 147 

Period  of   a  simple  harmonic 

motion 228 

of  a  wave  motion    ....  238 

Phase 229 

Phase  constant  ...  228,  233,  239 

Pitiscus 192 

Plane  sailing 118 

Plane,  surveying      ....   73^  118 

Plumb  line 72 


264 


INDEX 


Plutarch 7 

Points  of  the  compass      ...  75 
Pothenot's  problem      ....  123 
Principal  values  of  inverse  func- 
tions   219 

Products  of  sines  and  cosines    .  187 

Profile  of  a  wave 236 

Projection,  definition  of  .     .     .  201 

theorems  on 207 

of  a  broken  line      ....  208 

Proportional  parts,  tables  of     .  53 

Protractor 3 

Ptolemy 87,  192 

Pythagoras,  generalized  theo- 
rem of 85 

Quadrantal  formulae     .     .      171-173 

Quadrants,  the  four     ....  145 

Radian 165 

Radius,  of  circumscribed  circle  96 

of  escribed  circle     ....  97 

of  inscribed  circle   ....  94 

Radius  vector 139 

Ratios,  trigonometric  functions 

defined  as 15 

Reflection  of  light 128 

Refraction  of  light       ....  129 
Relations,  between  functions  of 

a  single  angle       .     .     .18,  148 
between  functions  of   com- 
plementary angles    .     .  20,  169 
between  functions  of  supple- 
mentary angles    ....  89 
between  functions  of  sym- 
metric angles 167 

Rheticus    . 192 

Right  triangles,  of  unfavorable 

dimensions 80 

solution  by  logarithms    .     .  67 
solution    by  natural    func- 
tions      ........  29 


S,  the  auxiliary 
Secant  curve  .     . 


.    61,  198 
.     .     163 


Secant,  definition  of  ...     .       14 

graph  of .163 

line  representation  of      .     .     154 

Sector,  area  of 166 

Segment,  area  of 166 

Sense  of  a  line  segment  .     .     .     202 

Signs  of  coordinates  ....     138 

of  trigonometric  functions  .     145 

Sine,  definition  of .     .     .  14,  83,  140 

graph  of 161 

line  representation  of      .     .     152 

origin  of  name 155 

Sine  curve 161 

Sines,  law  of      .     .     .       84,  96,  212 

Sin  (a  ±  P) 184,  186 

Sin  2  a      . 189 

Sin  ^  ±  Sin  5 100 

Skeleton  form  of  a  computation      69 

Slide  rule 62 

Slope 12 

Small  angles,  functions  of  .     .       60 

Smoley's  tables 31 

Snellius 125,  130,  252 

Solution  of  oblique  triangles   107-131 
of  right  triangles    ...     29,  67 

Spirit  level 5 

Squares,  table  of 30 

Standard  position  of  an  angle  .     136 

Stratton 243 

Subtended  angle 74 

Subtraction  formulae  ....  186 
Sum  of  two  angles,  functions 

of 184,  185,  212 

Sum  of  two  sines  or  cosines      .     188 
Supplementary    angles,    func- 
tions of 89 

Symmetric  angles,  functions  of    167 

T,  the  auxiliary  ....  61,  198 
Tables,  explanation  of  .  25,  30,  57 
Tangent,  definition  of     .     .    14,  140 

graph  of 163 

line  representation  of       .     .     153 
Tangent  curve 163 


I 


INDEX 


265 


Tangents,  law  of 101 

Tan  (a  ±  P) 185,  186 

Thacher's  slide  rule     ....  65 

Thales  of  Miletus 7 

Theodolite 4 

Transit 4 

Trigonometric  functions,  defini- 
tions of  ...     .       14,  89,  140 
table  of  logarithms  of     .     .  57 
table  of  values  of    ...     .  25 
Trigonometric  interpolation .     .  243 
Trigonometry,  object  of  .     .     .  10 
Trough  of  a  wave 232 

Variation  of  functions      .     .     .  146 


Vector 127 

Velocities 127 

Velocity  of  propagation  of    a 

wave 236 

Vernier 5 

Vertical     lines,     planes     and 

angles 72,73 

Vieta 87,  101 

Von  Opel 105 

Wave  length 232 

Wave  motion      ......    235 

Zero,  division  by 142 


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